Continuity

Definition (Point)

Let $A\subset X$ and $f:A\to Y$. Let $p\in A$. We say $f$ is continuous at $p$ if $\forall \epsilon >0,\exists \delta >0$ such that $\forall x\in A$ with $d_X(x,p)<\delta$ we have $d_Y(f(x),f(y))<\epsilon$

$$(\forall \epsilon >0)(\exists \delta >0)(\forall x\in A,0<d_X(x,p)<\delta )(d_Y(f(x),f(p))<\epsilon )$$

Proposition (Isolated implies Continuity)

If $p$ is an isolated point of $A$, then any function $f:A\to Y$ is continuous at $p$.

Proof: $p$ is an isolated point $A$ means that $\exists {\delta }_0>0$ such that $\forall x\in A,d(x,p)<{\delta }_0⟹x=p$. Indeed, for every neighborhood of $p$, it does intersect with $A$.

Since no neighborhoods intersect, there is some “minimal distance” ${\delta }_0$ where there are no points from $0$ to ${\delta }_0$. So, the only point must be itself.

Let $\epsilon >0$. Choose $\delta ={\delta }_0$. Then we have

$$d(f(x),f(p))=d(f(p),f(p))=0<\epsilon$$

and so it is continuous.

Definition (Set)

Let $A\subset X,f:A\to Y$. Then $f$ is continuous on $A$ if $\forall p\in A$, $f$ is continuous at $p$.

Examples:

1. Every polynomial $f(z)={\sum }_{k=0}^{\mathbb{C}}{a}_k{z}^k$ is continuous on $\mathbb{C}$ when $a_k\in \mathbb{C}$.
2. Let $f:X\to {\mathbb{R}}^n,f(x)=(f_1(x),f_2(x),\dots ,f_n(x))$ where $f_i:X\to \mathbb{R}$ for $i=1,\dots ,n$. Then $X$ is continuous on $X$ iff $f_i$ is continuous on $X$ for $i=1,\dots ,n$.

Lemma (Continuity of Composition)

Let $A\subset X,f:A\to Y$. Let $g:f(A)\to Z$. Define $h:=g\circ f:A\to Z$, where $h(a)=g(f(a))$.

Proof: Suppose $(a_n)\subset A,a_n\to p$. WTS $h(a_n)\to h(p)$.

1. Note since $a_n\to p$ and $f$ is continuous at $p$, then $f(a_n)\to f(p)$. Let $b_n:=f(a_n)$. Note $b_n\in f(A)=\text{dom}(g)$.
2. Since $b_n\to f(p)$ by $(1.)$ and $g$ is continuous at $f(p)$ then then $g(b_n)\to g(f(p))$. Thus $h(a_n)=g(f(a_n))=g(b_n)\to g(f_p)=h(p)$ by $(2.)$.

Proposition (Topological Characterization of Continuity)

Let $f:X\to Y$. Then $f$ is continuous on $X$ iff $\forall$ open sets $O\subset Y$, the preimage $f^{-1}(O)$ is open in $X$.

Proof: $(⟹)$ Suppose $f$ is continuous on $X$ and let $O$ be an open set in $Y$.

The goal is that $f^{-1}(O)$ is open in $X$. That is, $\forall x\in f^{-1}(O)$, we need to find $r>0$ such that $N_r(x)\subset f^{-1}(O)$.

1. Since $x\in f^{-1}(O)$ then $f(x)\in O$.
2. Since $O$ is open, $\exists \epsilon >0$ such that $N_{\epsilon }(f(x))\subset O$.
3. Since $f$ is continuous at $X$, $\exists \delta >0$ such that $\forall z\in X,d(x,z)<\delta ⟹d(f(x),f(z))<\epsilon$. That is, $z\in N_{\delta }(x)⟹f(z)\in N_{\epsilon }(f(x))$. So, $f(N_{\delta }(x))\subset N_{\epsilon }(f(x))\subset O$. Let $r=\delta$ to get $N_r(x)\subset f^{-1}(O)$. So $x$ is an interior point of $f^{-1}(O)$ and $f^{-1}(O)$ is open.

$(⟸)$ Assumption: for every $O$ open in $Y$, then $f^{-1}(O)$ is open in $X$.

Goal, $f$ is continuous at every $x\in X$.

Let $x\in X$. Let $\epsilon >0$. $N_{\epsilon }(f(x))$ is open in $Y$. So, by assumption, $f^{-1}(N_{\epsilon }(f(x)))$ is open in $X$. Note $x\in f^{-1}(N_{\epsilon }(f(x)))$. So $x$ is an interior point.

This implies $\exists \delta >0$ such that $N_{\delta }(x)\subset f^{-1}(N_{\epsilon }(f(x)))$. This implies $\forall z\in N_{\delta }(x)$, we have $f(z)\subset N_{\epsilon }(f(x))$.

So $\forall z$ with $d(x,z)<\delta$, we have $d(f(x),f(z))<\epsilon$.

Proposition (Continuity Retains Compactness)

Let $f:X\to Y$ be continuous. Let $K\subset X$ be compact. Then $f(K)$ is compact.

Proof: To show $f(K)$ is compact, let $\{O_{\alpha }\}$ be an open cover of $f(K)$.

Goal is to show there exists a finite subcover.

Since we have a cover,

$$f(K)\subset \bigcup _{\alpha \in I}{O}_{\alpha }$$

Take the preimage of this containment. Then

$$\begin{array}{rl}{f}^{-1}(f(K))& \subset f^{-1}\left(\bigcup _{\alpha \in I}{O}_{\alpha }\right)\\ & =\bigcup _{\alpha \in I}{f}^{-1}(O_{\alpha })\end{array}$$

Note $K=f^{-1}(f(K))$. So

$$K\subset \bigcup _{\alpha \in I}{f}^{-1}(O_{\alpha })$$

Since $f$ is continuous and $O_{\alpha }$ is open, then $f^{-1}(O_{\alpha })$ is open by lemma. So $\{f^{-1}(O_{\alpha })\mid \alpha \in I\}$ is an open cover of $K$. Since $K$ is compact, $\exists {\alpha }_1,\dots {\alpha }_n$ such that

$$K\subset f^{-1}(O_{{\alpha }_1})\cup \cdots \cup f^{-1}(O_{{\alpha }_n})$$

Apply $f$ to this containment,

$$\begin{array}{rl}f(K)& \subset f(f^{-1}(O_{{\alpha }_1}))\cup \cdots f(f^{-1}(O_{{\alpha }_n}))\\ & \subset O_{{\alpha }_1}\cup \cdots \cup O_{{\alpha }_n}\end{array}$$

Thus $f(K)$ has a finite subcover and is compact by definition.

Corollary (Extreme Value Theorem)

Let $X$ be a metric space that is compact. Suppose $f:X\to \mathbb{R}$ is continuous. Then $f$ attains its $\max$ and $\min$. That is, $\exists p,q\in X$ such that

$$\begin{array}{rl}f(p)& =\sup \{f(x)\mid x\in X\}=\max \{f(x)\mid x\in X\}\\ f(q)& =\inf \{f(x)\mid x\in X\}=\min \{f(x)\mid x\in X\}\end{array}$$

In particular, if $f:[a,b]\to \mathbb{R}$ is continuous, then $f$ has a $\max ,\min$.

Proof: For the second claim, note $[a,b]\subset \mathbb{R}$ is compact by Heine-Borel Theorem so it follows from the first.

Note that $f(X)\subset \mathbb{R}$ is compact by theorem and so by Heine-Borel Theorem $f(X)$ is closed and bounded. Then,

$$\begin{array}{r}\sup (f(X))\in \overline{f(X)}=f(X)\\ \inf (f(X))\in \overline{f(X)}=f(X)\end{array}$$

So $\exists p,q\in X$ such that $f(p)=\sup f(X)$ and $f(q)=\inf f(X)$. Note that $\sup ,\inf$ are in the closure by Lemma (Bounded).

Definition (Uniformly Continuous)

Let $f:X\to Y$. We say $f$ is Uniformly Continuous if

$$(\forall \epsilon >0)(\exists \delta >0)(\forall p,q\in X,d(p,q)<\delta )⟹(d(f(p),f(q))<\epsilon )$$

• Note that compared to the definition of continuity, we are saying there is a $\delta$ that “works” for all points in $X$.
• In particular, uniformly continuous implies continuous.

Example 1

Consider $f:\mathbb{R}\to \mathbb{R}$ with $f(x)=x^2$. We saw $f$ is continuous. However, $f$ is not uniformly continuous.

Proof: Assume by contradiction that $f$ is uniformly continuous. Let $\epsilon =1$. Then by definition, $\exists \delta >0$ such that $\forall x,y\in \mathbb{R},|x-y|<\delta$ and so $|f(x)-f(g)|<1$.

Fix $x\in \mathbb{R},x>0$. Let $y=x+\delta /2$. Then $(x-y)=|x-(x+\delta /2)|=\delta /2<\delta$. So by uniform continuity,

$$|x^2-y^2|=|x^2-(x+\delta /2{)}^2|=|x^2-(x^2+\delta x+{\delta }^2/4)|=|\delta x+{\delta }^2/4|=\delta x+{\delta }^2/4<1$$

Let $x=1/\delta$. Then

$$\delta x+{\delta }^2/4=1+{\delta }^2/4>1$$

which is a contradiction.

Example 2

$f:\mathbb{R}\to \mathbb{R}$ where $x\mapsto x$ is uniformly continuous.

Example 3

$f:\mathbb{R}\to \mathbb{R}$ where $x\mapsto \sqrt{x}$ is uniformly continuous.

Proof: Let $\epsilon >0$. Let $\delta ={\epsilon }^2$.

$$\begin{array}{rl}|f(x)-f(y){|}^2& =|\sqrt{x}-\sqrt{y}{|}^2\\ & =|\sqrt{x}-\sqrt{y}|\cdot |\sqrt{x}-\sqrt{y}|\\ & \le |\sqrt{x}-\sqrt{y}|\cdot (\sqrt{x}-\sqrt{y})\\ & =|\sqrt{x}-\sqrt{y}|\cdot |\sqrt{x}+\sqrt{y}|\\ & =|x-y|\\ & <\delta \\ & ={\epsilon }^2\end{array}$$

Thus, $|f(x)-f(y)|<\epsilon$ .

Example 4

$f:[-10,10]\to \mathbb{R}$ where $x\mapsto x^2$ is uniformly continuous.

Proof: Let $\epsilon >0$. Then let $\delta =\epsilon /20$. Suppose $|x-y|<\delta$ for $x,y\in [-10,10]$. Then

$$\begin{array}{rl}|f(x)-f(y)|& =|x^2-y^2|\\ & =|x-y|\cdot |x+y|\\ & <\delta (|x|+|y|)\\ & =\delta \cdot 20\\ & <\epsilon \end{array}$$

Done.

Theorem (Continuity + Compact = Uniform Continuity)

Let $X$ be a compact metric space. Let $f:X\to Y$ continuous. Then $f$ is uniformly continuous.

Proof: Suppose by contradiction that $X$ is not uniformly continuous. Then

$$(\exists {\epsilon }_0>0)(\forall \delta =1/n,\forall n\in \mathbb{N})(\exists p_n,q_n\in X)(d(p_n,q_n)<1/n)\text{ but }d(f(p_n),f(q_n))\ge {\epsilon }_0$$

Now since $X$ is compact, $(p_n)\subset X$, there $\exists$ convergent subsequence $(p_{n_k}),p_{n_k}\to p$.

I claim $q_{n_k}\to p$. Proof:
This is $\alpha /2$ argument. Let $\alpha >0$. Choose $K$ large enough so that

$$d(p_{n_k},p)<\alpha /2$$

and

$$d(p_{n_k},q_{n_k})<1/n_k<\alpha /2$$

for $k\ge K$ which we can do since $p_{n_k}\to p$ and by premise + Archimedean Property. Then $\forall k\ge K$, $d(q_{n_k},p)\le d(q_{n_k},p_{n_k})+d(p_{n_k},p)<\alpha$ and so they converge to the same point.

Since $f$ is continuous on $X$ and $p\in X$. Then since $p_{n_k}\to p$, then

$$\underset{k\to \infty }{\lim }f(p_{n_k})=f(p)$$

by Lemma (Sequential Characterization of Limits). Indeed,

$$q_{n_k}\to p⟹f(q_{n_k})\to f(p)$$

So for $K$ large, $d(f(p_{n_k}),f(q_{n_k}))<{\epsilon }_0$. This is a contradiction.

Lemma (Cauchy Test for Uniform Continuity)

Suppose $f:X\to Y$ is uniformly continuous. Then for every Cauchy Sequence $(p_n)$ in $X$, we have $(f(p_n))\subset Y$ is Cauchy.

Proof: Let $\epsilon >0$. Since $f$ is uniformly continuous, $\exists \delta >0$ such that

$$\forall p,q\in X,d(p,q)<\delta ⟹d(f(p),f(q))<\epsilon$$

Given that $(p_n)\subset X$ is Cauchy, $\exists N\in \mathbb{N}$ such that $\forall n,m\ge N,d(p_n,p_m)<\delta$. So since $f$ is uniformly continuous, $\forall n,m\ge N,d(f(p_n),f(p_m))<\epsilon$ and so $(f(p_n))$ is Cauchy.

Example 5

$f:(0,1]\to \mathbb{R}$ for $x\mapsto 1/x$ is not uniformly continuous.

Proof: Note $(1/n)$ is Cauchy. So suppose by contradiction that $f$ is uniformly continuous. Then by lemma, $(f(1/n))$ is Cauchy. But $f(1/n)=n$ which is not Cauchy (the sequence becomes $(n)$), a contradiction.

Proposition (Noncompactness Properties)

Let $A\subset \mathbb{R}$ be not compact.

1. There exists a continuous function on $A$ which is not bounded.
2. There exists a bounded continuous function that has no max on $A$.
3. If $A$ is bounded, then $\exists$ a continuous function on $A$ which is not uniformly continuous. In general, boundedness is necessary in $(3)$. Indeed if $A=\mathbb{N}$, then every function on $A$ is uniformly continuous because $\forall \epsilon >0,\delta =1/2$, so $\forall x,y\in A$, if $|x-y|<\delta$ then $x=y$ so $|f(x)-f(y)|<\epsilon$.

Proof: Consider the case where $A$ is bounded but not compact. Since $A\subset \mathbb{R}$, by Heine-Borel Theorem, $A$ cannot be closed.

Then $\exists b\in \mathbb{R}$ such that $b$ is a limit point, but $b\notin A$.

$(1)$ Consider $f(x)=1/(x-b)$. Since $b\notin A$, $f$ is defined and it is continuous on $A$ by algebra of continuous functions and by Example 2.

• But $f$ is not bounded. (Prove this)
• $f$ is not uniformly continuous. (Prove this)

$(2)$ Let

$$g(x)=\frac{1}{1+(x-b{)}^2}$$

Note $1+(x-b{)}^2>1$. So $g(x)<1,\forall x\in A$. However since $\exists x_n\to b$ we see $\sup \{g(x)\mid x\in A\}$. It is bounded, but max.

$(3)$ Exercise.

Theorem (Continuity Retains Connectedness)

If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, and if $C$ is a connected subset of $X$, then $f(C)$ is connected.

Proof: Suppose by contradiction that $f(C)$ is separated. Then $f(C)=A\cup B$ where both sets are nonempty. Then, we get

$$\begin{array}{rl}f(C)& =A\cup B\\ C& =f^{-1}(A\cup B)\\ & =f^{-1}(A)\cup f^{-1}(B)\end{array}$$

Let $E=C\cap f^{-1}(A)$ and $F=C\cap f^{-1}(B)$. Note that $C=E\cup F$. Both $E,F$ are empty; otherwise, $f(C)\subset B$, and $A=\varnothing$ a contradiction.

Since $B\subset \overline{B}$, then $f^{-1}(B)\subset f^{-1}(\overline{B})$. As $f$ is continuous and $\overline{B}$ is closed, by theorem, we have that $f^{-1}(\overline{B})$ is closed. So

$$\overline{f^{-1}(B)}\subset \overline{f^{-1}(\overline{B})}=f^{-1}(\overline{B})$$

Then, $F=C\cap f^{-1}(B)⟹F\subset f^{-1}(B)$. Thus,

$$\overline{F}\subset \overline{f^{-1}(B)}\subset f^{-1}(\overline{B})$$

OTOH, $E=C\cap f^{-1}(A)\subset f^{-1}(A)$. Indeed,

$$E\cap \overline{F}\subset f^{-1}(A)\cap f^{-1}(\overline{B})=\varnothing$$

and similarly for $\overline{E}\cap F$. Thus, $E,F$ are separated, a contradiction, as $C$ is connected.

Theorem (Continuity + Compact = sup/inf)

Suppose $f$ is a continuous real function on a compact metric space $X$, and

$$M=\underset{p\in X}{\sup }f(p)m=\underset{p\in X}{\inf }f(p)$$

Then there exists $p,q\in X$ such that $f(p)=M$ and $f(q)=m$. The proof uses Heine-Borel Theorem.

Intermediate Value Theorem

Let $f$ be a real valued continuous function on $[a,b]$. If for $r$,

$$f(a)<r<f(b)$$

then $\exists c\in [a,b]$ such that $f(c)=r$.

Proof: WLOG, suppose $f(a)<f(b)$. Let $r\in [f(a),f(b)]$. WLOG, suppose $r\in (f(a),f(b))$. By the theorem, $f([a,b])\subset \mathbb{R}$ is connected. Let $I=f([a,b])$. Since $I$ is an interval, $f(a)\in I,f(b)\in I$ and $f(a)<r<f(b)$. then $r\in I=f([a,b])⟹r=f(c)$ for some $c\in [a,b]$.

Definition (Directional Limits)

Let $f:(a,b)\to \mathbb{R}$. Let $p\in (a,b)$. We say the limit from the right at $p$ equals $\ell$ and write

$$\underset{x\to p^{+}}{\lim }f(x)=\ell$$

if

$$(\forall \epsilon >0)(\exists \delta >0)(x\in (a,b),0<x-p<\delta )⟹(|f(x)-\ell |<\epsilon )$$

Equivalently, we say ${\lim }_{x\to p^{+}f(x)}=\ell$ if $\forall (x_n)$ in $(p,b)$ such that $x_n\to p$ we have $f(x_n)\to \ell$. This is denoted as $f(p+)$ and $f(p-)$ for the limit from the left.

Remark (Equivalent Limit Direction)

1. $f$ has a limit at $p$ $⟺$ $f(p+)=f(p-)$. In particular, both exist.
2. $f$ is continuous at $p$ $⟺$ $f(p+)=f(p-)=f(p)$.

Definition (Discontinuity)

Let $f:(a,b)\to \mathbb{R}$. We say $f$ has a discontinuity of the first kind at $p$ if $f$ is discontinuous at $p$ but $f(p+)$ and $f(p-)$ both exist.

Otherwise, we say $f$ has a discontinuity of the second kind.