Generalizations

Generalizations in ${\mathbb{R}}^n$ §

We first generalize many of the concepts we learned about in MATH410 to higher dimensions.

Vector Spaces §

$R^n$ is a vector space, with vectors and scalars (respectively) given as

$$u=(u_1,u_2,\dots u_n)\in {\mathbb{R}}^n,u_i\in \mathbb{R}$$

Satisfying vector addition and scalar multiplication, and inner product commonly known as the dot product

$$<u,v>=u_1{v}_1+\cdots +u_n{v}_n$$

Satisfying the following axioms of a (real) inner product space:

• $<u,v>=<v,u>$
• $<\alpha u+\beta v,w>=\alpha <u,w>+\beta <v,w>$
• $<u,u>\ge 0$, and $<u,u>=0$ if and only if $u=0$.
• $||u|{|}^2=<u,u>$

Theorem: Inner Product in ${\mathbb{R}}^2$

In ${\mathbb{R}}^2$, $<u,v>=||u||\cdot ||v||\cos \theta$ where $\theta$ is the angle between the two vectors.

Proof

Using polar coordinates, we can convert our vectors into the polar space

$$u=||u||(\cos \alpha ,\sin \alpha )v=||v||(\cos \beta ,\sin \beta )$$

Where $\alpha$ is the angle of $u$ to the $x$-axis, and $\beta$ is the angle of $v$ to the same axis.

We can find that

$$<u,v>=||u||||v||(\cos \alpha \cos \beta +\sin \alpha \sin \beta )=||u||||v||\cos (\beta -\alpha )$$

We say that $u$ is orthogonal to $v$, $u\perp v$, if $<u,v>=0$.

Theorem: Properties of Orthogonality

$$||u+v|{|}^2=||u|{|}^2+||v|{|}^2$$

If and only if $⟨u,v⟩=0$.

Proof

$$\begin{array}{rl}||u+v|{|}^2& =⟨u+v,u+v⟩=⟨u,u⟩+⟨u,v⟩+⟨v,u⟩+⟨v,v⟩\\ & =||u|{|}^2+2⟨u,v⟩+||v|{|}^2\end{array}$$

Which is equal to $||u|{|}^2+||v|{|}^2$ if and only if $⟨u,v⟩=0$.

Cauchy-Schwarz Inequality

Let $u,v\in {\mathbb{R}}^n$. Then,

$$|⟨u,v⟩|\le ||u||\cdot ||v||$$

We prove this using the axioms of inner products only (see above), making generalizable for any inner product!

Proof

First, note that if either $u$ or $v$’s normals are 0, then this is trivially true.

Assume that both $||u||\ne 0$ and $||v||\ne 0$. Now, take the expression $||u-\alpha v|{|}^2$, and note that this is (by an axiom) $\ge 0$.

By a previous proof,

$$||u-\alpha v|{|}^2=||u|{|}^2-2\alpha ⟨u,v⟩+{\alpha }^2||v|{|}^2=||u|{|}^2+\alpha (-2⟨u,v⟩+\alpha ||v|{|}^2)$$

We seek to minimize this with respect to $\alpha$. Ignoring the constants, we minimize

$$2⟨u,v⟩+\alpha ||v|{|}^2=0\to \alpha =\frac{⟨u,v⟩}{||v|{|}^2}$$

Subbing this in, we have

$$\begin{array}{rl}0& \le ||u-\alpha v|{|}^2=||u|{|}^2-2\frac{⟨u,v⟩}{||v|{|}^2}⟨u,v⟩+{\left(\frac{⟨u,v⟩}{||v|{|}^2}\right)}^2||v|{|}^2=||u|{|}^2-\frac{⟨u,v{⟩}^2}{||v|{|}^2}\\ 0& \le ||u|{|}^2-\frac{⟨u,v{⟩}^2}{||v|{|}^2}\\ ⟨u,v{⟩}^2& \le ||u|{|}^2||v|{|}^2\\ |⟨u,v⟩|& \le ||u||||v||\end{array}$$

Example: Cauchy-Schwarz (1)

Find the maximum value of

$$\frac{x+2y}{\sqrt{x^2+y^2}}$$

Let ${\vec{v}}_1=(x,y)$, ${\vec{v}}_2=(1,2)$. Then, by Cauchy-Schwarz,

$$\begin{array}{rl}|{\vec{v}}_1\cdot {\vec{v}}_2|& \le ||{\vec{v}}_1||\cdot ||{\vec{v}}_2||\\ x+2y& \le \sqrt{x^2+y^2}\sqrt{5}\\ \frac{x+2y}{\sqrt{x^2+y^2}}& \le \sqrt{5}\end{array}$$

We find our maximium is $\sqrt{5}$! Now, we find our maximizer.

Cauchy-Schwarz is only equal when ${\vec{v}}_1={\vec{v}}_2$. So,

$$x=1y=2$$

We have maximizer $(1,2)$!

Example: Cauchy-Schwarz (2)

Find the maximum value of

$$\frac{x+y}{\sqrt{x^2+4y^2}}$$

Let ${\vec{v}}_1=(x,2y)$, and ${\vec{v}}_2=(1,1/2)$. Then, by Cauchy Schwarz,

$$\begin{array}{rl}|{\vec{v}}_1\cdot {\vec{v}}_2|& \le ||{\vec{v}}_1||\cdot ||{\vec{v}}_2||\\ x+y& \le \sqrt{x^2+4y^2}\sqrt{5/4}\\ \frac{x+y}{\sqrt{x^2+4y^2}}& \le \sqrt{5/4}\end{array}$$

We have found a maximum value! Now, we find our maximizer.

For Cauchy-Schwarz to be equal, it must be true that ${\vec{v}}_1={\vec{v}}_2$. So,

$$x=12y=1/2⟹y=1/4$$

Our maximizer is $(1,1/4)$!

Corollary

Note that equality holds if and only if $\exists \alpha \in \mathbb{R}$ such that $u=\alpha v$ (or $v=\alpha u$).

Proof ( $\leftarrow$)

$$u=\alpha v\to |⟨u,v⟩|=|\alpha |||v|{|}^2=||\alpha v||||v||=||u||||v||$$

Proof ( $\to$)

Let $||u||||v||=|⟨u,v⟩$. Then, if we repeat our original proof, we find that

$$||u|{|}^2-\frac{⟨u,v{⟩}^2}{||v|{|}^2}=\frac{||u|{|}^2||v|{|}^2-⟨u,v{⟩}^2}{||v|{|}^2}=0$$

Telling us that $||u-\alpha v||=0$ for our minimizer $\alpha =\frac{⟨u,v⟩}{||v|{|}^2}$. So, $u-\alpha v=0$ by an axiom.

Theorem: Triangle Inequality

$$||u+v||\le ||u||+||v||\forall u,v\in {\mathbb{R}}^n$$

Proof

We find that

$$||u+v|{|}^2=||u|{|}^2+2⟨u,v⟩+||v|{|}^2=||u|{|}^2+2||u|{|}^2||v|{|}^2+||v|{|}^2$$

Giving us our triangle inequality due to the Cauchy-Schwarz Inequality.

Theorem: Reverse Triangle Inequality

$$||u\pm v||\le \left|||u||-||v||\right|$$

Proof ( $-$)

$$||u||=||u-v+v||\le ||u-v||+||v||$$

We also know that

$$||u-v||\ge ||u||-||v||$$

Giving us

$$||v-u||\le ||v||-||u||$$

Sequences and Convergence §

Consider a sequence in ${\mathbb{R}}^n$, denoted $\{u_k\}$ for all $k\in \mathbb{N}$. In other words, each entry in the sequence is a point in ${\mathbb{R}}^n$.

Note that by convention we will denote $u^i$ as the projection of the $i^{th}$ element of $u$ ($u$’s $i^{th}$ element).

If $\{u_k\}$ is a sequence in ${\mathbb{R}}^n$, and for some $u\in {\mathbb{R}}^n$ the following holds:

$$\underset{k\to \infty }{\lim }{u}_k=u\text{if}\underset{k\to \infty }{\lim }||u_k-u||=0$$

Then we call this the limit of the sequence, alternatively written as $\{u_k\}\to u$ ($u_k$ converges to $u$) in the norm.

This is similar in idea to convergence in 1-dimensional series!

But in ${\mathbb{R}}^n$, this is not the only notion of convergence that we have! We can also say that $\{u_k\}\to u$ componentwise, if each $u_k^i\to u^i$ as $k\to \infty$.

Theorem: Synonymous Notions of Convergence

$\{u_k\}\to u$ in norm if and only if $\{u_k\}\to u$ componentwise.

Proof

Proof ($\to$) §

Assume that $\{u_k\}\to u$ in norm. Then, for each $1\le i\le n$,

$$0\le |u_k^i-u^i|\le \sqrt{(u_k^1-u^1{)}^2+\cdots +(u_k^n-u^n{)}^2}=||u_k-u||\to 0$$

So by the squeeze theorem, $|u_k^i-u^i|\to 0$.

Proof ($\leftarrow$) §

Assume that $\{u_k^i\}\to u^i$. Then,

$$||u_k-u||=\sqrt{(u_k^1-u^1{)}^2+\cdots +(u_k^n-u^n{)}^2}$$

As each component converges to 0 by definition, and sums of sequences converging to 0 also converge to 0, we know that $||u_k-u||\to 0$.

Note that this does not hold if we have infinite dimensions ($n$ not fixed)! In particular, component-wise convergence does not imply convergence in norm.

Remark: Dissimilarity of Convergence in Infinite Dimensions

Say we have a sequence $\{u_k\}$, where for any $k$ the $k^{th}$ element is 1, all other 0. So,

$$\begin{array}{rl}{u}_1& =(1,0,0\dots )\\ u_2& =(0,1,0,\dots )\\ u_3& =(0,0,1,\dots )\end{array}$$

We can see that as $k\to \infty$, $\{u_k\}\to \vec{0}$, the 0 vector! However, $||u_k||\to 1$, so it does not converge in norm to 0.

Closed and Open Sets §

Let $r>0$, $u\in {\mathbb{R}}^n$. We define the open ball of radius $r$, centered at $u$ as

$$B_r(u)=\{v\in {\mathbb{R}}^n:||u-v||<\mathbb{R}\}$$

In other words, the set of all $n$-dimensional points that are within some predetermined hypersphere of $u$.

In $\mathbb{R}$, this is the open interval around a value $u$!

We say that a set $A\subseteq {\mathbb{R}}^n$ is open if $\forall u\in A$, $\exists r>0$ such that $B_r(u)\subseteq A$. In other words, for all vectors in $A$, we can find an open ball around the vector such that all vectors in the ball are in $A$.

Theorem: Open Balls and Open Sets

For $r>0$, $u\in {\mathbb{R}}^n$, the open ball $B_r(u)$ is an open set.

Intuitively, this makes sense as for any vector inside the open ball, we can find a ball contained within our original ball. As our vectors get closer and closer to the ball’s edge, our containing ball will shrink!

Proof

We want to show that $\forall v\in B_r(u)$, $\exists p>0$ such that $B_p(v)\le B_r(u)$.

We know that for $v$, $||u-v||\mathbb{R}$ by definition. We define the distance from $v$ to the ball’s edge, $p=r-||u-v||$. This will be the radius of our new ball around $v$.

We now show that $B_p(v)\subseteq B_r(u)$. Pick any $w\in B_p(v)$. By definition,

$$||v-w||<p⟹||v-w||<r-||u-v||$$

Look at $||u-w||$. We know that

$$\begin{array}{rl}||u-w||& =||u-v+v-w||\\ & \le ||u-v||+||v-w||\\ & <||u-v||+p\\ & <||u-v||+(r-||u-v||)\\ & <r\end{array}$$

Thus, $w\in B_r(u)$!

We say that a set $A\subseteq {\mathbb{R}}^n$ is closed if for any sequence in $A$, $\{u_k\}\subseteq A$, such that $\{u_k\}\to u$, $u\in {\mathbb{R}}^n$, then $u\in A$. In other words, for any convergent sequence in $A$, the vector it converges to is also in $A$.

Furthermore, if $A\in {\mathbb{R}}^n$, then we call $A^c$ the complement of $A$, the set of all vectors not in $A$.

$$A^c={\mathbb{R}}^n\backslash A=\{u\in {\mathbb{R}}^n|u\notin A\}$$

Theorem: Open Sets and their Complements

$A$ is an open set if and only if its complement, $A^c$, is closed.

Proof

Proof ($\to$) §

Assume first that $A$ is open. To show that $A^c$ is closed, pick any sequence $\{u_k\}\subseteq A^c$ such that $\{u_k\}\to u$, $u\in {\mathbb{R}}^n$.

We want to show that $u\in A^c$. Suppose $u\notin A^c$, so $u\in A$. Then, because $A$ is open, there exists an $r>0$ such that

$$B_r(u)\subseteq A$$

But no $u_k$ can be in this ball $B_r(u)$! This violates an assumption, so $u$ cannot be in $A$.

Proof ($\leftarrow$) §

Conversely, assume $A^c$ is closed. To show that $A$ is open, choose any $u\in A$. By definition, $\exists r>0$ such that $B_r(u)\subseteq A$.

By way of contradiction, assume that $\forall r>0$, our ball around $u$ is not a subset of $A$. This means that $B_r(u)$ contains some $v\notin A$ ($v\in A^c$), where $v$ depends on $r$.

Let $r=1/a$. For all $a\in \mathbb{Z}$, we can form a sequence of $v_a$’s which, by assumption, are all in $A^c$.

$$||v_a-u||<r{v}_a\in A^c$$

However, as $a\to \infty$, $r\to 0$! This tells us that $\{v_a\}\to u,u\in A$, violating our assumption that $A^c$ is closed.

Note that the statement “$A$ is open if and only if it is not closed” is not true. There exist sets that are neither open or closed, and there exist sets that are both open and closed (ex. the empty set and ${\mathbb{R}}^n$)!

We continue by discussing intersections and unions of open and closed sets.

Intersections / Unions of Open Sets

Let $\{V_s{\}}_s$ be a (possibly infinite) family of open sets. Then, the union of these open sets ${\bigcup }_{s\in S}{V}_s$ is also open.

Let $V_1,\dots V_k$ be a finite family of open sets. Then, the intersection of these sets $V_1\cap \cdots \cap V_k$ is also open.

Note that the intersection clause only works for finite sets. We can find a counterexample for an infinite family by choosing open balls with radius $r=1/a$, so that as $a\to \infty$, we are left with a point.

Proof: Intersections and Unions of Open Sets

Union of Open Sets §

Pick $u\in {\bigcup }_{s\in S}{V}_s$. We know that $\exists s_0\in S$ such that $u\in V_{s_0}$. Since $V_{s_0}$ is open, we know that $\exists r>0$ such that $B_r(u)\in V_{s_0}\subseteq {\bigcup }_{s\in S}{V}_s$.

Intersection of Open Sets §

Let $u\in V_1\cap \dots V_k$. We know that by definition,

• Because $u\in V_1$, $\exists r_1>0$ such that the ball $B_{r_1}(u)\subseteq V_1$
• Because $u\in V_2$, $\exists r_2>0$ such that $B_{r_2}(u)\subseteq V_2$
• $\dots$

Continue this argument for all $k$. Now let $r=\min (r_1,\dots r_k)$. Then $B_r(u)\subseteq B_{r_i}u$ for all $i=1\dots k$, meaning $B_r(u)\subseteq {\bigcap }_{i=1}^k{V}_i$.

Intersections / Unions of Closed Sets

Let $F_1,\dots F_k$ be a finite family of closed sets. Then their union $F_1\cup \dots F_k$ is closed.

Let $\{F_s{\}}_s\in S$ be a possibly infinite family of closed sets. Then, their intersection ${\bigcap }_{s\in S}{F}_s$ is closed.

Proof: Intersections and Unions of Closed Sets

Intersection Proof (1) §

We know that by Demorgan’s Law, ${\left({\bigcap }_{s\in S}{F}_s\right)}^c={\bigcup }_{s\in S}{F}_s^c$. We also know from an earlier theorem that each $F_s^c$ is open, and any union of open sets is open. So ${\left({\bigcap }_{s\in S}{F}_s\right)}^c$ is open, and therefore ${\bigcap }_{s\in S}{F}_s$ is closed.

Intersection Proof (2) §

Let us have a sequence $\{u_k\}\subseteq {\bigcap }_{s\in S}{F}_s$, and assume $\{u_k\}\to u\in {\mathbb{R}}^n$. By definition of intersection, we know that $u_k\in F_s$ for all $s\in S$! So, by definition of a closed set, $u\in F_s$ for all $s\in S$.

This means that $u\in {\bigcap }_{s\in S}{F}_s$.

Union Proof (1) §

We know that by Demorgan’s Law, ${\left(F_1\cup \dots F_k\right)}^c=F_1^c\cap \dots F_k^c$. Each set is, by an earlier theorem, open.

As etablished previously, the intersection of open sets yields an open set, so ${\left(F_1\cup \dots F_k\right)}^c$ is open. By an earlier theorem, this means that $F_1\cup \dots F_k$ is closed.

Union Proof (2) §

Let us define a sequence in this union, $\{u_i\}\subseteq F_1\cup \dots F_k$, where $\{u_i\}\to u\in {\mathbb{R}}^n$.

Because we have infinitely many $u_i$’s, and a finitely many sets $F$, at least one of these sets must contain infinitely many $u_i$ by the Pidgeonhole Principle. Let this set be $F_j$.

So, there exists a subsequence $\{u_{i_l}{\}}_{l\in \mathbb{N}}$ contained in $F_j$. As $\{u_i\}\to u$, it must be true that this subsequence converges to $u$ as well, and as $F_j$ is closed (by definition), $u\in F_j$. So, $u\in F_1\cup \dots F_k$.

Using the concept of an open ball, we can formally define the interior, exterior, and boundary of a set. Let $A\subseteq {\mathbb{R}}^n$.

We define the interior of $A$ as

$$\text{int}A=\{u\in {\mathbb{R}}^n:\exists r>0\text{s.t.}{B}_r(u)\subseteq A\}\subseteq A$$

In other words, the set of all points that can be contained within a ball inside of $A$.

We can also consider the interior of $A$’s complement. By definition, these two sets are mutually exclusive.

But what’s between them? We call this the boundary:

$$\text{Boundary}A=\{u\in {\mathbb{R}}^n:\forall r>0,B_r(u)\text{meets both}A\text{and}{A}^c\}$$

These 3 sets are mutually exclusive, and form all of $R^n$!

• The interior of $A$
• The interior of $A^c$
• The boundary of $A$

Continuity and Compactness §

Continuity of Functions §

Let $A\subseteq {\mathbb{R}}^n$ be a set, and define the function $F:A\to {\mathbb{R}}^m$.

We say $F$ is continuous at $u\in A$ if $\forall \{u_k\}\to u$ where $\{u_k\}\subseteq A$, it follows that $\{F(u_k)\}\to F(u)$. In other words, for any sequence in $A$ converging to $u$, the sequence evaluated in the function $F(u_k)$ should converge to $F(u)$!

Example: Continuity

Let $f:\{1,2\}\cup [3,\infty )\to \mathbb{R}$ such that

$$f(x)=\{\begin{array}{ll}1& x=1\\ 0& x=2\\ x& \text{else}\end{array}$$

Is $f$ continuous at $x=1$? Yes! If $\{u_k\}\subseteq A$ such that $\{u_k\}\to u$, then $u_k=1$ for $k$ sufficiently large.

Let $ϵ=\frac{1}{10}$. Then, by definition of convergence, $\exists k$ such that

$$|u_k-1|<\frac{1}{10}\forall k\ge K$$

But $\{u_k\}\subset \{1,2\}\cup [3,\infty )$! Thus, $u_k=1$ $\forall k\ge K$, meaning $f(u_k)=f(1)=1$.

Theorem: Preservation of Continuity

Let $f,g:A\to \mathbb{R}$ such that $A\subseteq {\mathbb{R}}^n$. If $f,g$ are continuous at $u\in A$, then

• $\forall \alpha ,\beta \in \mathbb{R}$, $\alpha f+\beta g$ are also continuous at $u$

• If $\forall v\in A,g(v)\ne 0$, then $\frac{f}{g}$ is continuous at $u$.

Proof

Let $\{u_k\}\to u$, and assume $\{u_k\}\subseteq A$.

Then, by assumption, $\{f(u_k)\}\to f(u)$, and $\{g(u_k)\}\to g(u)$, so

$$\begin{array}{rl}& (f+g)(u_k)\to (f+g)(u)\\ & fg(u_k)\to fg(u)\end{array}$$

By properties of sequences of real numbers.

Similarly,

$$\frac{f(u_k)}{g(u_k)}\to \frac{f(u)}{g(u)}$$

if all $g(v)\ngtr 0$.

Theorem: Compositions of Functions

Suppose we have two sets $A\subseteq {\mathbb{R}}^n$, $B\subseteq {\mathbb{R}}^m$, and define functions

$$\begin{array}{r}F:A\to B\\ G:B\to \mathbb{R}\end{array}$$

If $F$ is continuous at $u\in A$, and $G$ is continuous at $F(u)$, then the composition $G\circ F$ is continuous at $u$.

Proof

Let $\{u_k\}\to u$. We know that $\{F(u_k)\}\to F(u)$ by definition and $\{G(F(u_k))\}\to G(F(u))\}$, i.e. $\{(G\circ F)(u_k)\}\to G\circ F(u)$.

We can also prove continuity using what’s known as an $ϵ$-$\delta$ definition.

By $ϵ$-$\delta$, $F:A\to {\mathbb{R}}^m$, $A\subseteq {\mathbb{R}}^n$ is continuous at $u\in A$ if and only if $\forall ϵ>0$, $\exists \delta >0$ such that

$$||u-v||<\delta \to ||F(u)-F(v)||<ϵ$$

Proof (Epsilon-Delta Continuity)

Proof ($\leftarrow$) §

Assume $ϵ$-$\delta$ definition.

Let $\{u_k\}\to u$. Let $ϵ>0$. We want $\exists K$ such that $\forall k\ge K$,

$$||F(u_k)-F(u)||<ϵ$$

Because $\{u_k\}\to u$, we know by defintion that $\exists K$ such that $\forall k\ge K$,

$$||u_k-u||<\delta$$

Then

$$||F(u_k)-F(u)||<ϵ\forall k\ge K$$

Proof ($\to$) §

Assume that the sequence definition of continuity holds. Assume by contradiction, that the $ϵ$-$\delta$ definition does not hold.

So, $\exists ϵ>0$ where $\forall \delta >0$, $\exists v\in A$ such that

$$||u-v||<\delta \to ||F(u)-F(v)||\ge ϵ$$

Let $\delta =\frac{1}{k}$, and $\forall k\in \mathbb{N}$ form a sequence with these $v$‘s. This sequence $\{v_k\}\to u$! However, by assumption, for all $v_k$,

$$||F(v_k)-F(u)||\ge ϵ$$

So the sequence $\{F(v_k)\}$ does not converge to $F(u)$, which is a contradiction!

We also find the following equivalence definitions for epsilon-delta. First, recall that if $F:A\to {\mathbb{R}}^m$, then for $B\subseteq A$, we define

$$F(B)=\{F(b):b\in B\}$$

And if $C\subseteq {\mathbb{R}}^m$, then

$$F^{-1}(C)=\{u\in A:F(u)\in C\}$$

Or in other words, the set of all inputs to $F$ yielding possible outputs in $C$.

Note that this does not require the function inverse to exist. We’re just defining the set of all inputs in $A$ yielding something in $C$.

Using these definitions, we claim that the epsilon-delta deinition is also equivalent to saying $\forall ϵ>0$, $\exists \delta >0$,

1. $$F(B_{\delta }(u)\cap A)\subseteq B_{ϵ}(F(u))$$

   In other words, all inputs of the function in a $\delta$ ball centered around $u$ must have an output which is within the set of all possible outputs of the function in the $ϵ$ ball centered around $u$.

   Note that we intersect with $A$ to guarantee that we have an input for the function, as not all vectors in a ball around $u$ may be an input for the function.

2. $$B_{\delta }(u)\cap A\subseteq F^{-1}(B_{ϵ}(F(u)))$$

   In other words, all possible inputs within the $\delta$ ball of $u$ are contained within the set of inputs that yield the $ϵ$ ball of $F$ around $u$.

Theorem: Continuity and Open Sets

Let $O$ be open in ${\mathbb{R}}^n$, and let $F:O\to {\mathbb{R}}^m$.

Then, $F$ is continuous at every point of $O$ if and only if the set of inputs $F^{-1}(V)$ is open for all $V\subseteq {\mathbb{R}}^m$ open.

Proof

Proof $\to$ §

Let $F$ be continuous for all $x\in O$. We wish to show that the set of inputs $F^{-1}(V)$ is open $\forall V\in {\mathbb{R}}^n$.

Let $V\in {\mathbb{R}}^n$. We want to show that $F^{-1}(V)$ is open. In other words, $\forall x\in F^{-1}(V)$, $\exists r>0$ such that $B_r(x)\subseteq F^{-1}(V)$.

Because $x\in F^{-1}(V)$, $F(x)\in V$. By the openness on $V$, we can find a $r>0$ such that

$$B_r(F(x))\subseteq V$$

Because all values of $B_r(F(x))$ are in $V$, all values of the domain mapping to values $B_r(F(x))$ also map to values in $V$. Thus, taking the inverse $F^{-1}$ maintains the subset property.

$$F^{-1}(B_r(F(x)))\subseteq F^{-1}(V)$$

Because $F$ is continuous, we find that $\forall ϵ>0$, $\exists \delta >0$

$$B_{\delta }(x)\cap O\subseteq F^{-1}(B_{ϵ}(F(x)))$$

We let $ϵ=r$, to find a ${\delta }_1>0$ such that

$$B_{{\delta }_1}(x)\cap O\subseteq F^{-1}(B_r(F(x)))\subseteq F^{-1}(V)$$

We now end by modifying our $\delta$ to guarantee that $B_{\delta }(x)\subseteq O$. Because $x\in F^{-1}(V)$, it is in the domain of $F$ ($O$), and as $O$ is open, we can find a ${\delta }_2>0$ such that $B_{{\delta }_2}\subseteq O$. Choose $\delta =\min ({\delta }_1,{\delta }_2)$. Then, we find tha

$$B_{\delta }(x)\cap O=B_{\delta }(x)\subseteq F^{-1}(B_r(F(x)))\subseteq F^{-1}(V)$$

Thus, $F^{-1}(V)$ is open.

Proof ($\leftarrow$) §

Suppose that for all $V\subseteq {\mathbb{R}}^m$ open, we have that $F^{-1}(V)$ is also open. We want to show that $F$ is continuous.

Let $x\in O$. Now let $ϵ>0$. We want to find a $\delta >0$ such that

$$F(B_{\delta }(x)\cap O)\subseteq B_{ϵ}(F(x))$$

Take the ball $B_{ϵ}(F(x))$. Because open balls are always open, and $B_{ϵ}(F(x))\subseteq {\mathbb{R}}^m$, we know that $F^{-1}(B_{ϵ}(F(x)))$ is open by assumption.

Because this is open, and $x\in F^{-1}(B_{ϵ}(F(x))$, by definition, we can find a $r>0$ such that

$$B_r(x)\subseteq F^{-1}(B_{ϵ}(F(x))$$

Let $\delta =r$. We have found our $\delta$!

$$B_r(x)\cap O\subseteq B_r(x)\subseteq F^{-1}(B_{ϵ}(F(x))$$

We have a similar case with closed sets.

Theorem: Continuity and Closed Sets

Let $F:{\mathbb{R}}^n\to {\mathbb{R}}^m$.

Then, $F$ is continuous at every point of $O$ if and only if the set of inputs $F^{-1}(V)$ is closed for all $V\subseteq {\mathbb{R}}^m$ closed.

Proof

Proof ($\to$) §

Let $F$ be continuous. We want to show that $F^{-1}(V)$ is closed for all $V\subseteq {\mathbb{R}}^m$ closed.

Choose some $V\subseteq {\mathbb{R}}^m$ closed, and take its complement, $V^c$ which is open. Because $V^c$ is open and $F$ is continuous, by the previous theorem it must hold that $F^{-1}(V^c)$ is open.

Note that if $x\in F^{-1}(V^c)$, then it cannot be in $F^{-1}(V)$, as $V^c\cap V=\varnothing$ and a single domain value cannot map to two values. Thus, $F^{-1}(V^c)=F^{-1}(V{)}^c$! As $F^{-1}(V{)}^c$ is open, it therefore must hold that $F^{-1}(V)$ is closed.

Proof ($\leftarrow$) §

Suppose that for all $V\subseteq {\mathbb{R}}^m$ closed, $F^{-1}(V)$ is closed. We wish to show that $F$ is continuous.

We show that if for any $V\subseteq {\mathbb{R}}^m$ open, $F^{-1}(V)$ is open, then $F$ is continuous. Let $V\subseteq {\mathbb{R}}^m$ open. Then, $V^c$ is closed, so by assumption, $F^{-1}(V^c)$ is closed. By a similar argument from before, $F^{-1}(V^c)=F^{-1}(V{)}^c$, so $F^{-1}(V)$ is open.

Remark

Let $f:{\mathbb{R}}^n\to \mathbb{R}$. If $f$ is continuous, $\alpha \in \mathbb{R}$ fixed, then the following sets are open.

$$\begin{array}{r}{f}^{-1}((-\infty ,a))=\{u\in {\mathbb{R}}^n:f(u)<\alpha \}\\ f^{-1}((\alpha ,\infty ))=\{u:{\mathbb{R}}^n:f(u)>\alpha \}\end{array}$$

Similarly, $\{u\in {\mathbb{R}}^n:f(u)\le \alpha \}$ is closed because $\{u\in {\mathbb{R}}^n:f(u)\le \alpha {\}}^c=\{u\in {\mathbb{R}}^n:f(u)>\alpha \}$ is open.

Consider the following examples.

Example: Continuity and Closure

Let $f:\mathbb{R}\to \mathbb{R}$ continuous. Prove that

$$G=\{(x,f(x)):x\in \mathbb{R}\}$$

is closed.

Let $\{a_k\}\in G$, where $\{a_k\}\to a$. We wish to show that $a\in G$.

By definition of $G$, we can express our sequence $\{a_k\}$ as

$$\{a_k\}=\{(x_k,f(x_k))\}\to (x,y)$$

Because $\{x_k\}\to x$, by continuity on $f$ we know that

$$\{f(x_k)\}\to f(x)$$

And by the uniqueness of limits, if $\{f(x_k)\}\to f(x),\{f(x_k)\}\to y$, then $y=f(x)$. Thus,

$$\{x_k,f(x_k)\}\to (x,f(x))\in G$$

Thus, $G$ is closed.

Let’s now consider the converse. Let $f:\mathbb{R}\to \mathbb{R}$. Assume $G=\{(x,f(x)):x\in \mathbb{R}\}$ is closed. Does it follow that $f$ is continuous?

Let $\{x_k\}\in \mathbb{R}$, where $\{x_k\}\to x$. We want to show that $\{f(x_k)\}\to f(x)$.

Using this, we can form a sequence $\{(x_k,f(x_k))\}\subseteq G$. Then, if $\{(x_k,f(x_k)\}\to (x,y)$, then $f(x)=y$. So, if there exists a $y$ such that $\{f(x_k)\}\to y$, then $y=f(x)$ where $\{x_k\}\to x$.

But do we know if $\{f(x_k)\}$ converges? In fact, closure only says that if the sequence converges, then its result is in the set! However, we may be able to find a function that fails to satisfy this.

Let

$$f(x)=\frac{1}{x},f(0)=0x_k=\frac{1}{k}$$

Then, $f(x)$ is not continuous at $x=0$, yet the set $G$ is still closed!

Sequential Compactness §

A set $A\subseteq {\mathbb{R}}^n$ is bounded if $\exists M$ such that

$$||u||\le M\forall u\in A$$

In other words, there exists an $M$ which is an upper bound for all vector lengths in the set.

We say set $K\subseteq {\mathbb{R}}^n$ is sequentially compact if for all sequences $\{u_l\}\subseteq K$, there exists a subsequence $\{u_{l_m}\}$ and $u\in K$ such that

$$\{u_{l_m}\}\to u$$

In other words, all sequences have a convergent subsequence that converge to something in $K$.

We show that sequential compactness is synonymous with closure and boundedness (together).

Theorem: Sequential Compactness and Boundedness

If $K\subseteq {\mathbb{R}}^n$ is sequentially compact, then $K$ is bounded.

Proof

We want $\exists M\ge 0$ such that

$$||u||\le Mu\in K$$

By way of contradiction, assume that $K$ is sequentially compact but not bounded.

Then, $\forall M>0$, $\exists u\in K$ such that

$$||u||>M$$

Now let $M=m\in \mathbb{N}$, to get a sequence $u_m\in K$ such that $||u_m||>M$. For any subsequence in this sequence,

$$||m_{m_l}||>m_l\to \infty$$

So $u_{m_l}$ cannot converge! This is a contradiction.

Theorem: Sequential Compactness and Closure

If $K\subseteq {\mathbb{R}}^n$ is sequentially compact, then $K$ is closed.

Proof

Pick a sequence of points $\{u_l\}\subseteq K$, and assume that this sequence $\{u_l\}\to u\in {\mathbb{R}}^n$. We want to show that $u\in K$.

By sequential compactness, there exists a subsequence $\{u_{l_m}\}\to v$ such that $v\in K$. But $\{u_l\}\to u$, so $\{u_{l_m}\}\to u$! As limits are unique, this implies $u=v$, so $u\in K$!

Recall that if we have a sequence $\{x_l\}\subseteq \mathbb{R}$, and $\{x_l\}$ is bounded, then there exists a subsequence $\{x_{l_m}\}$ and $x\in \mathbb{R}$ such that

$$\{x_{l_m}\}\to x$$

This is known as the Bolzano-Weirstrass Theorem. We can generalize this to ${\mathbb{R}}^n$.

Bolzano Weierstrass Theorem (Higher Dimensions)

If $\{u_l\}\subseteq {\mathbb{R}}^n$, and $\{u_l\}$ is bounded, then $\exists \{u_{l_m}\}\to u\in {\mathbb{R}}^n$.

Proof by Induction

We know that if $n=1$, this is true.

Assume true for $n$. We wish to prove that this theorem holds for $n+1$.

Let $\{u_l\}\in {\mathbb{R}}^{n+1}$, bounded. Write

$$u_l=(u_l^{\prime },u_l^{n+1})$$

Where $u_l\in {\mathbb{R}}^{n+1}$, $u_l^{\prime }\in {\mathbb{R}}^n$, $u_l^{n+1}\in \mathbb{R}$.

By assumption, $\exists \{u_{l_m}^{\prime }\}\to u^{\prime }\in {\mathbb{R}}^n$. Now look at the corresponding sequence of $u^{n+1}$, $\{u_{l_m}^{n+1}\}$. By our base case, $\exists \{u_{l_{m_p}}^{k+1}\}\to u^{n+1}\in \mathbb{R}$. Then,

$$\{u_{l_{m_p}}\}\to (u^{\prime },u^{n+1})$$

Theorem: Closed + Bounded and Compactness

$K\subseteq {\mathbb{R}}^n$ is sequentially compact if and only if $K$ is closed and bounded.

Proof

We already proved $\to$. We now prove the converse.

Let $K$ be closed and bounded. Choose any sequence $\{u_l\}\subseteq K$. As $K$ is bounded, by the Bolzano-Weierstrass Theorem we know that there exists a subsequence $\{u_{l_m}\}\to v$, where $v\in {\mathbb{R}}^n$.

But we also know that $K$ is closed! So, as $\{u_{l_m}\}\subseteq K$, it must be true that $v\in K$. So,

$$\exists \{u_{l_m}\}\to v\in K$$

We find that functions on sequentially compact sets have some guaranteed properties.

Theorem: Images of Sequentially Compact Domains

If $K$ is sequentially compact and $F:K\to {\mathbb{R}}^m$ is continuous, then the image, $F(K)$, is sequentially compact.

Proof

Let $\{v_l\}\subseteq F(K)$. We want to show that there is a subsequence that converges to something in $F(K)$.

By definition of the image, we know that $\exists \{u_k\}\subseteq K$ such that

$$F(u_k)=v_k$$

By sequential compactness on $K$, there is a subsequence $\{u_{k_l}\}\to u\in K$, and furthermore, $\{F(u_{k_l})\}\to F(u)$ as $F$ is continuous.

Note that both closure and boundedness (implying compactness) must hold for this to be true! They do not hold on their own.

Example: Failure of Image Closure and of Images

If $C\subseteq {\mathbb{R}}^n$ is closed, $F:C\to {\mathbb{R}}^m$ is continuous, is $F(C)$ closed?

No! As a counterexample, let $f:\mathbb{R}\to \mathbb{R}$, where $f=\frac{1}{1+x^2}$. Then, $\mathbb{R}$ is closed, but $f$ is not, as it asymptotes towards $0\notin f(\mathbb{R})$!

If $B\subseteq \mathbb{R}$ is bounded, and $f:B\to \mathbb{R}$ is continuous, is $f(B)$ bounded?

No! Let $f(x)=\frac{1}{x}$ on $B=(0,1)$. Then, $F(B)=(1,\infty )$ which is unbounded.

Note the specific wording on the domain.

If $B\subseteq \mathbb{R}$ is bounded and $f:\mathbb{R}\to \mathbb{R}$ is continuous (domain is now $\mathbb{R}$), now $f(B)$ is bounded! We can find a sequentially compact domain containing $B$, and find that $f$ on this domain is sequentially compact (and thus bounded!).

We also find that functions on sequentially compact sets obey the extreme value theorem.

Theorem: Extreme Value Theorem (Generalization)

Let $K$ be sequentially compact, $f:K\to \mathbb{R}$ continuous.

Then $f$ has a minimum and maximum value.

Proof

We know that from an earlier theorem, $f(K)$ is bounded.

By definition, $\exists m$ which is the infimum of $f$ over all $K$. We want to show that $\exists u\in K$ such that $f(u)=m$.

By definition of infimum, we know that $\exists \{u_i\}\subseteq K$ such that $\{f(u_i)\}\to m$. Since $K$ is sequentially compact, we know there exists a convergent subsequence $\{u_{i_l}\}$ such that $\{u_{i_l}\}\to u\in K$. Then, by continuity,

$$\lim f(u_{i_l})=\lim f(u_i)=m$$

Interestingly enough, the converse of the above theorem can actually be used to prove sequential compactness.

Theorem: Sequential Compactness via Minimums and Maximums

Let $A\subseteq {\mathbb{R}}^n$ be a set such that any function $f:A\to \mathbb{R}$ has a minimum and maximum value. Then, $A$ is sequentially compact.

Proof

We wish to show that $A$ is bounded and closed.

To show that $A$ is bounded, choose some function $f(x)=||x||$. By assumption, we know that $\exists x_{\text{max}}$ such that $\forall x\in A$,

$$f(x)\le f(x_{\text{max}})⟹0\le ||x||\le ||x_{\text{max}}||$$

Let $x_{\text{max}}$ be our bound.

To show that $A$ is closed, let $\{u_i\}\in A$ such that $\{u_i\}\to u\in {\mathbb{R}}^n$. We want to show that $u\in A$.

Let $f(x)=||x-u||$. Note that $f$’s infimum is 0. Thus, the minimum of $f$ is also 0! But the minimum can only be attained at $x=u$, so $u\in A$.

We also say set $K\subseteq {\mathbb{R}}^n$ is compact if for every family of open sets $\{V_{\alpha }\}$ such that $K\subseteq {\bigcup }_{\alpha }{V}_{\alpha }$, then there exist finitely many sets $V_{{\alpha }_1},\dots ,V_{{\alpha }_l}$ such that

$$K\subseteq V_{{\alpha }_1}\cup \cdots \cup V_{{\alpha }_l}$$

In other words, $K$ is compact if for any family of (possibly infinite) open sets covering $K$, we can find a finite number of sets that cover $K$.

If $K\subseteq {\bigcup }_{\alpha }{V}_{\alpha }$, we say that ${\bigcup }_{\alpha }{V}_{\alpha }$ covers $K$.

We similarly prove compactness with respect to closure and boundedness.

Theorem: Compactness vs. Closure and Boundedness

$K$ is compact if and only if $K$ is closed and bounded.

Proposition 1: $K$ compact implies $K$ bounded.

Let $K\subseteq {\bigcup }_{l=1}^{\infty }{B}_l(0)$ (this is always true, as this union is all of ${\mathbb{R}}^n$). Then by definition, there must exist some “largest” ball that contains $K$ given by a finite value. So, there is a $l_0$ such that $K\subseteq B_{l_0}$.

So, $K$ is bounded.

Proposition 2: $K$ compact implies $K$ closed.

Let $K$ be compact, and let $\{u_k\}$ be a sequence of points in $K$, where $\{u_k\}\to u\in {\mathbb{R}}^n$. We wish to show that $u\in K$.

Assume by contradiction that $u\notin K$.

Now, we must make a family of open sets whose union is guaranteed to cover (contain) $K$, so we can apply our definition.

Let $V_l=\{x\in {\mathbb{R}}^n:||x-u||>1/l\}$. Then, the union of all $V_l$’s is all of ${\mathbb{R}}^n$, with the exception of the point $u$.

We are looking at the set of points outside of circles that are closing in on $u$.

$$\bigcup _l{V}_l=\{x\in {\mathbb{R}}^n:x\ne u\}$$

Therefore $K$ must be contained in this family of open sets. By definition, there must exist a finite amount of sets that contains $K$. But since each of these sets contain each other, their union is the same as the largest set, meaning we can find a $V_{l_0}$ such that

$$K\subseteq V_{l_0}$$

This contradicts $\{u_k\}\subseteq K$, $\{u_k\}\to u$, as there cannot be any $\{u_k\}$ in the ball $V_{l_0}$!

We can choose an epsilon to show that convergence fails, as there is an open ball around $u$ that $K$ is NOT a part of.

Proposition 3: If the set $L$ is compact, and $K\subseteq L$, $K$ is closed, then $K$ is compact.

Let $V_{\alpha }$ be a family of open sets where $K\subseteq {\bigcup }_{\alpha }{V}_{\alpha }$. We wish to find a finite amount of sets in this family containing $K$.

For any $x$, if it is in $K$, then it is in ${\bigcup }_{\alpha }{V}_{\alpha }$. Otherwise, if must be in $K^c$. Because $K$ is closed, $K^c$ must be open. So,

$$L\subseteq \bigcup _{\alpha }{V}_{\alpha }\cup K^c$$

Since $L$ is compact, there must exist a finite amount of sets $V_{{\alpha }_1},V_{{\alpha }_2},\dots ,V_{{\alpha }_l}$ from this family such that

$$L\subseteq \bigcup _{i=1}^l{V}_{{\alpha }_i}$$

As $K\subseteq L$, then,

$$K\subseteq \bigcup _{i=1}^l{V}_{{\alpha }_i}$$

But this union could contain $K^c$! So, we show that $K^c$ cannot be in this union, to find a finite amount of sets from $\{V_{\alpha }\}$ covering $K$.

By definition, a point in $K$ cannot be in $K^c$! So, $K$ is contained in a finite number of the $V_{\alpha }$.

Proof ( $\leftarrow$)

Note that this proof heavily relies on the fact that $K$ is in finitely many dimensions.

Let $K$ be closed and bounded. Since $K$ is bounded, $K$ is contained in some closed cube in ${\mathbb{R}}^n$. Without loss of generality, assume $K\subseteq C=[0,1]\times [0,1]\cdots \times [0,1]$.

It suffices to show that this cube is compact, after which we apply the previous theorem (because $K$ is closed) to get our result.

Let $\{V_{\alpha }\}$ be open sets such that $C\subseteq {\bigcup }_{\alpha }{V}_{\alpha }$. Assume by contradiction that we cannot find a finite number of sets from $\{V_{\alpha }\}$ containing $C$.

Divide $C$ into $2^n$ subcubes of size $\frac{1}{2}$. We show that these subcubes can be contained in finitely many sets, and therefore, so can $C$.

By assumption, we must be able to find a closed subcube $C_1$ which cannot be covered by finitely many $V_{\alpha }$‘s. Continue, dividing $C_1$ into $2^n$ subcubes of size $\frac{1}{4}$. Then, among these subcubes, there must be one that cannot be covered by finitely many sets. Repeating this, we get

$$C_i\subseteq \cdots \subseteq C_2\subseteq C_1\subseteq C$$

Where no $C_i$ can be covered by finitely many sets.

By a generalization of the nested interval theorem, there must exist a single unique point contained within all nested cubes- formally, $\exists !x_0\in C_i\forall i$.

We know that as $x_0\in C$, the point must exist in at least one of the $V_{\alpha }$’s, say $V_{{\alpha }_0}$. By definition of an open set, there exists a ball around $x_0$ completely contained within $V_{{\alpha }_0}$.

So, we can find some $C_i\in V_{{\alpha }_0}$, which is covered by finitely many sets. This is a contradiction!

Example

Let $f:\mathbb{R}\to \mathbb{R}$. Assume that $f^{-1}(K)$ is closed for every (sequentially) compact set $K$ in $\mathbb{R}$. Is $f$ continuous?

Not necessarily. The theorem above requires that all closed sets are closed in $f^{-1}(K)$, but not all closed sets are sequentially compact.

Let $f$ be the function

$$f(x)=\{\begin{array}{ll}\frac{1}{x}& x\ne 0\\ 0& x=0\end{array}$$

Then, every sequentially compact set $K$ on $f$’s range maps to a closed set in $f$’s domain, but $f$ is not continuous at 0!

Uniform Continuity §

The function $F:A\to {\mathbb{R}}^n$, $A\subseteq {\mathbb{R}}^m$ is uniformly continuous if for any two sequences $\{u_i\},\{v_i\}\subseteq A$, if

$$\{||u_i-v_i||\}\to 0$$

Then,

$$\{||f(u_i)-f(v_i)||\}\to 0$$

Note that uniform continuity implies continuity. Simply let $v_i$ be the sequence of whatever $\{u_i\}$ converges to so we get our continuity definition!

The converse is not true. See the example below.

Example: Continuity Doesn't Imply Uniform Continuity

Let $f(x)=x^2$, which is a continuous function.

Then, for $u_i=\frac{1}{i}+i$, $v_i=i$, we have that $u_i-v_i\to 0$, but $f(u_i)-f(v_i)=2+\frac{1}{i^2}\nrightarrow 0$!

Theorem: Uniform Continuity and Sequential Compactness

If $K$ is sequentially compact, and $F:K\to {\mathbb{R}}^n$ is continuous, then $F$ is uniformly continuous.

Proof

Let $K$ be sequentially compact, and $F:K\to {\mathbb{R}}^n$ continuous. We wish to show that $F$ is uniformly continuous.

By way of contradiction, suppose that $F$ is not uniformly continuous. Then, there exists some $\{x_k\},\{y_k\}$ such that $\{x_k-y_k\}\to 0$, but

$$\{f(x_k)-f(y_k)\}\nrightarrow 0$$

By definition of continuity, $\exists ϵ>0$ such that $\forall N\in \mathbb{N}$, $\forall k\ge N$,

$$||f(x_k)-f(y_k)||\ge ϵ$$

Choose this $ϵ$. Then, by sequential compactness of $K$, we can find a subsequence such that for $i\ge N$, all terms have a difference greater than $ϵ$.

$$||f(x_{k_i})-f(y_{k_i})||\ge ϵ$$

But by sequential compactness on each $f(x_{k_i})$, $f(y_{k_i})$, we can find convergent subsequences

$$\{f(x_{k_{i_j}})\}\to u_1\{f(y_{k_{i_j}})\}\to u_2$$

As $f$ is continuous, and $\{x_k\},\{y_k\}$ converge, then $u_1=u_2$. This means that our subsequence $f(x_{k_i})-f(y_{k_i})$ converges to 0, but this is a contradiction!

Note that there exist non-compact sets that also satisfy this property! Consider the following example.

Example

Let $C\subseteq {\mathbb{R}}^n$ with the property that any $f:C\to {\mathbb{R}}^n$ continuous is uniformly continuous. Is it true that $C$ must be closed and bounded?

Must $C$ be Closed? §

Let $\{x_k\}\in C$, where $\{x_k\}\to x_0$. If $C$ is bounded, then we need to show that $x_0\in C$.

By way of contradiction, suppose $x_0\notin C$. Then, we can define function

$$f(x)=\frac{1}{|x-x_0|}$$

Which is continuous on $C$, as $x=x_0$ is by assumption not continuous on $C$.

By assumption, $f(x)$ must be uniformly continuous. So, $\forall ϵ>0$, $\exists \delta >0$ such that $\forall x,y\in C$,

$$|x-y|<\delta \to |f(x)-f(y)|<ϵ$$

Let $ϵ=1$. By assumption, we have a delta $\delta >0$ such that for $|x-y|<\delta$, $|f(x)-f(y)|<1$ for all $x,y$ satisfying this condition.

Choose $x_k,y_k$ converging to $x_0$ such that

$$|x_k-x_0|<\frac{\delta }{2}|y_k-x_0|<\frac{\delta }{2}$$

Then,

$$|x_k-y_k|<\delta$$

Fix $x_k$, but let $y_k\to x_0$. Then, our assumptions should still hold as we are within $\delta$, but $f(y_k)\to \infty$! This is a contradiction.

So, $C$ must be closed.

Must $C$ be Bounded? §

Suppose $C=\mathbb{N}$.

Let $f:\mathbb{N}\to \mathbb{R}$ continuous. Then, $f$ is automatically continuous, as in the integers, any convergent sequence must hover at the same epsilon value (due to the spacing between the integers)- it’ll be the same integer value (due to the spacing between integers).

Let $ϵ>0$. Let $\delta =1/2$. If $x,y\in \mathbb{N}$, then if $|x-y|<1/2$, then $x=y$, so $f(x)=f(y)$, so $|f(x)-f(y)|=0<ϵ$.

However, $C$ is not bounded! So, $C$ does not have to be bounded.

Like continuity, we also have a $ϵ-\delta$ definition for uniform continuity too!

Let $F:A\to {\mathbb{R}}^m$. Then, the following are equivalent.

1. If $\{u_i\},\{v_i\}\subseteq A$ with $\{||u_i-v_i||\}\to 0$, then $\{||F(u_i)-F(v_i)||\}\to 0$.
2. $\forall ϵ>0$, $\exists \delta >0$ such that if $u,v\in A$ with $||u-v||<\delta$, then $||F(u)-F(v)||<ϵ$.

The latter is the $ϵ-\delta$ definition for uniform continuity.

Proof

Proof (1 to 2) §

We prove this by contrapositive. Assume that (2) is not true. We wish to show that as a result, (1) cannot be true.

By the negation of (2), $\exists ϵ>0$ such that $\forall \delta >0$, $\exists u,v\in A$, $||u-v||<\delta$ with $||F(u)-F(v)||\ge ϵ$.

Choose $\delta =\frac{1}{k}$, and let $k\in \mathbb{N}$ to create sequences $\{u_k\},\{v_k\}$ such that $||u_k-v_k||<\frac{1}{k}\to 0$, where $||F(u_k)-F(v_k)||\ge ϵ$.

Thus, (1) is not true.

Proof (2 to 1) §

Assume (2). Let $\{u_i\},\{v_i\}\subseteq A$ with $\{||u_i-v_i||\}\to 0$. We wish to show that $\{||F(u_i)-F(v_i)||\}\to 0$.

Let $ϵ>0$. Then, $\exists \delta >0$ such that (2) holds. Let $I$ such that $||u_i-v_i||<\delta$. Then, by (2), $\forall i\ge I$, $||F(u_i)-F(v_i)||<ϵ$.

Convexity and Connectedness §

We say the set $A$ is convex if $\forall u,v\in A$, $\forall 0\le t\le 1$,

$$t(u)+(1-t)v\in A$$

In other words, for any two points in the set, all points along the line between these two points are also within the set.

In lower dimensions, this defines a convex shape!

We say $A\subseteq {\mathbb{R}}^n$ is path connected if $\forall u,v\in A$, if there exists a continuous function $\gamma :[a,b]\to A$, with

$$\gamma (a)=u\gamma (b)=v$$

In other words, for any two points in the set, we can define a function (a “path”) between these two points that remains completely within the shape! We call $\gamma$ (gamma) the parameterized path.

In lower dimensions, this defines a completely connected shape!

Theorem: Path Connected Subsets of the Real Line

On the real line, $A\subseteq \mathbb{R}$ is path connected if and only if $A$ is an interval.

Proof

Proof ($\leftarrow$) §

If $A$ is an interval, let $u,v$ be points in this interval. Define function $\gamma (t)=tu+(1-t)v$ for $0\le t\le 1$. We can trivially show that $\gamma (0)=u,\gamma (1)=v$, and furthermore, $\gamma$ is continuous. We have our parameterized path.

Proof ($\to$) §

Conversely, if $A\subseteq \mathbb{R}$ is path connected, let $u,v\in A$. We wish to show that the interval $[u,v]\subseteq A$, which implies that $A$ is an interval (as we chose $u,v$ arbitrarily).

We know that $\exists \gamma :[a,b]\to A$ continuous with $\gamma [a]=u,\gamma [b]=v$. Choose any $w\in (u,v)$. We wish to show that $w\in A$, which then shows that $[u,v]\subseteq A$ ($u,v$ are trivially in $A$).

By the intermediate value theorem, $\exists c\in (a,b)$ such that $\gamma (c)=w$! Thus, $w\in A$ as it is in the image of $\gamma$, which is defined to be $A$.

Theorem: Path Connected Sets on Continuous Functions

If $A\subseteq {\mathbb{R}}^n$ is path connected and $F:A\to {\mathbb{R}}^m$ is continuous, then the image $F(A)$ is also path connected. In other words, continuity preserves path connectedness!

Proof

Let $u,v\in F(A)$. We wish to find a parameterized path between $u$ and $v$.

By definition of an image, we can find $x,y\in A$ such that $F(x)=u,F(y)=v$. By path connectedness of $A$, we can find an interval $[a,b]$ with a parameterized function $\gamma$ such that $\gamma (a)=x,\gamma (b)=y$.

Then, as $F$ is also continuous, we can find a composition of the functions $F\circ \gamma :[a,b]\to F(A)$, which is continuous such that $F\circ \gamma (a)=F(x)=u$, and $F\circ \gamma (b)=F(y)=v$. We have found a parameterized path from $u$ to $v$.

We say that the set $A$ has the Intermediate Value Property (IVP) if for any continuous function defined on the set, $f:A\to \mathbb{R}$, the image $f(A)$ is an interval.

It can be shown that if $A$ is path connected, it has the intermediate value property!

Set $A\subseteq {\mathbb{R}}^n$ is not connected if $\exists U,V$ open sets such that:

1. $U\cap A\ne \varnothing ,V\cap A\ne \varnothing$
2. $(U\cap A)\cap (V\cap A)=\varnothing$
3. $(U\cap A)\cup (V\cap A)=A$

In other words, we can find two disjoint open sets that make up $A$. If $U,V$ satisfythese conditions, then we say $U,V$ separate $A$.

Thus, by this definition, a set is connected if we cannot find two disjoint open sets that make up $A$.

Note that by this more intuitive definition, we can deduce that ${\mathbb{R}}^n$ is connected, as we cannot form 2 disjoint open sets by partitioning the space- there must be one with a boundary.

Theorem: Connected Sets and IVP

$A$ is connected if and only if $A$ has the Intermediate Value Property.

Proof

We want to show that if $A$ is not connected if and only if A does not have the intermediate value property. In other words, $\exists f:A\to \mathbb{R}$ continuous, with $f(A)$ not an interval.

Proof ($\to$) §

Assume that $A$ is not connected, and let $U,V$ open separate $A$.

Define

$$f(x)=\{\begin{array}{ll}1& x\in A\cap U\\ 0& x\in A\cap V\end{array}$$

Note the following:

• $f$ is defined $\forall x\in A$ because of property (3) of not connected
• The value $f(x)$ is unique because of (2).
• $f(A)=\{0,1\}$ because of (1), because there is at least one element of $U$ in $A$, and one element of $V$ in $A$.

To show $f$ is continuous at $x_0\in U\cap A$, we want to show that $\exists \delta >0$ such that $B_{\delta }(x_0)\subseteq U$. Then, if $x\in A$ and $||x-x_0||<\delta$, then $f(x)=f(x_0)$, so $|f(x)-f(x_0)|<ϵ$ trivially.

So we found $f:A\to \mathbb{R}$, continuous, where $f(A)$ is not on an interval.

Proof ($\leftarrow$). §

Assume $A$ does not have the intermediate value property. Let $f:A\to \mathbb{R}$ continuous, with $f(A)$ not an interval.

Then, $\exists c\in \mathbb{R}$ such that $c\notin f(A)$, $f(A)\cap (-\infty ,c)\ne \varnothing$, $f(A)\cap (c,\infty )\ne \varnothing$.

To construct $U,V$ separating $A$, let

$$\tilde{U}=f^{-1}((-\infty ,c)),\tilde{V}=f^{-1}((c,\infty ))$$

Then, $\tilde{U}\ne \varnothing ,\tilde{V}\ne \varnothing$.

• $\tilde{U}\cup \tilde{V}=A$ because we know that $c\notin f(A)$.
• $\tilde{U}\cap \tilde{V}=\varnothing$, as there are no points that are in both $(-\infty ,c)$ and $(c,\infty )$.

To finish, we will find $U,V$ open such that $\tilde{U}=U\cap A$, $\tilde{V}=V\cap A$.

Let $x\in \tilde{U}$. Then, $f(x)<c$. Since $f$ is continuous, $\exists r>0$ such that $\forall y\in B_r(x)\cap A$, $f(y)<c$. Let $U={\cup }_{x\in \tilde{U}}{B}_r(x)$.

Notice that $U\cap A=\tilde{U}$, so ${\bigcup }_{x\in \tilde{U}}(B_r(x)\cap A)=\tilde{U}$.

We can also find that if $A$ is path connected, then $A$ is connected.

Remark: Connected $\nrightarrow$ Path Connected

Note that the converse is not true. If $A$ is connected, it may not be path connected.

For example, let $A=\{0\}\times [-1,1]\cup \{(x,\sin (1/x)):0<x\le 1\}$, which oscillates as we get closer to the $y$ axis, but never touches! Thus, even if $A$ is connected, it is not path connected.

Thus, we end with the following graph of implications. Below, an arrow indicates that the source implies the destination.

graph LR
1[Path Connected];
2[Connected];
3[Intermediate Value Property];

1 -.-> 2 & 3;
2 -.-> 3;
3 -.-> 2;

We can use the idea of path connectedness to prove a remark earlier about sets that are both open and closed!

Corollary

IF $U\subseteq {\mathbb{R}}^n$ is both open and closed, then $U={\mathbb{R}}^n$, or $U=\varnothing$

Proof

We know that ${\mathbb{R}}^n$ is path connected, so $\mathbb{R}$ is connected.

Let $U$ be both open and closed. Let $V=U^c$. Because $U$ is open, $V$ is also open.

In this case, $U\cap V=\varnothing$, and $U\cup V={\mathbb{R}}^n$!

So, $U\ne \varnothing ,V\ne \varnothing$ is impossible! So, it must be true that either $U={\mathbb{R}}^n,V=\varnothing$, or $U=\varnothing ,V={\mathbb{R}}^n$.