ANA1_L20

To do:

• uniform continuity
• connectedness

Recall

Definition of open cover compactness.

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Uniform continuity

Recall the definition of a continuous function:

Definition 1(Definition).

Let $f:X\to Y$. $f$ is continuous on $X$ if

$$\forall x\in X,\forall ϵ>0,\exists \delta >0\text{ such that }f(B_{\delta }(x,X))\subset B_{ϵ}(f(x),Y).$$

Compare with definition of uniform continuity:

Definition 2(Definition).

Let $f:X\to Y$. $f$ is uniformly continuous on $X$ if

$$\forall ϵ>0,\exists \delta >0\text{ such that }\forall x\in X,f(B_{\delta }(x,X))\subset B_{ϵ}(f(x),Y).$$

Things to note:

• Uniform continuity implies continuity, but the converse is generally not true. Consider $f:(0,\infty )\to \mathbb{R}$, $f(x)=\frac{1}{x}$ as a counterexample. The converse is true, if $X$ is compact, as we will prove.
• Continuity is a local phenomenon; you only need to look in a (potentially very tiny) neighborhood of $x$ and $f(x)$ to check if an $ϵ$ challenge can be met. Uniform continuity requires you to analyze the function over its entire domain.
• The definition for uniform continuity can be rephrased like so:
  $f:X\to Y$ is uniformly continuous if $\forall ϵ>0$, $\exists \delta >0$ such that if for some $p,q\in X$, $d_X(p,q)<\delta$ then $d_Y(f(p),f(q))<ϵ$.

Continuous functions defined on compact sets are uniformly continuous

Rudin, 4.19

Theorem 3(Theorem).

Let $X$ be compact, $f:X\to Y$ be continuous. Then, $f$ is uniformly continuous.

Proof
Let $ϵ>0$ be arbitrary. To show that $f$ is uniformly continuous, we have to produce a $\delta$ such that $d_X(p,q)<\delta$ implies $d_Y(f(p),f(q))<ϵ$. Since $f$ is continuous, we can find ${\delta }_x$ for every $x\in X$ such that $f(B_{{\delta }_x}(x))\subset B_{\frac{ϵ}{2}}(f(x))$. Note that the set

$$S\equiv \left\{B_{\frac{{\delta }_x}{2}}(x)|x\in X\right\}$$

is an open cover of $X$. Since $X$ is compact, $S$ must have a finite subcover

$$S^{\prime }\equiv \left\{B_{\frac{{\delta }_x}{2}}(x)|x\in K\subset X\right\}.$$

Let $\delta \equiv \min \left\{\frac{{\delta }_x}{2}|x\in K\right\}$. Consider $p$, $q$ in $X$ with $d_X(p,q)<\delta$. Now, $p$ must be in $B_{\frac{{\delta }_x}{2}}(x)$ for some $x\in K$. Thus, $d_X(p,x)<\frac{{\delta }_x}{2}$. Triangle inequality gives us $d_X(q,x)<{\delta }_x$. Thus, $p,q\in B_{{\delta }_x}(x)$. This implies $f(p),f(q)\in B_{\frac{ϵ}{2}}(f(x))$, i.e, $d_Y(f(p),f(q))<ϵ$. ❏

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Theorems that fail when X is not compact

Let $f:X\to \mathbb{R}$ be a continuous function. Let $X$ be compact. The extreme value theorem tells us that $f(x)$ attains a maximum and minimum value, and the previous theorem tells us that $f$ is uniformly continuous.

We will now show that these theorems do not hold when $X$ is not compact.

We know from the Heine Borel theorem that in $\mathbb{R}$, compact $⟺$ closed and bounded.
Also, if $f$ does not attain a maximum or minimum, it is either unbounded, or is bounded and $f(X)$ does not contain its supremum and infimum.

Rudin, 4.20

Theorem 4(Theorem).

Let $X\subset \mathbb{R}$ be non compact.

1. $\exists$ continuous unbounded $f:X\to \mathbb{R}$.
2. $\exists$ continuous bounded $f:X\to \mathbb{R}$ with no maximum value.
3. If $X$ is bounded, $\exists$ continuous $f:X\to \mathbb{R}$ that is not uniformly continuous

Note that the statement 3 with $X$ being unbounded is false. For example, take $X=\mathbb{Z}$. Any function defined on $X$ is continuous, and also uniformly continuous (take $\delta =1$).

Proof of 1
Note that in all the following examples $f$ is continuous since restrictions of continuous functions are continuous.
$X$ is not closed, $f$ is not bounded: $f(x)=\frac{1}{x-x_0}$, where $x_0\notin X$ and $x_0$ is a limit point of $X$.
$X$ is not bounded, $f$ is not bounded: $f(x)=x$. ❏

Proof of 2
$X$ is not closed, $f(X)$ does not contain its supremum/infimum: $f(x)=\frac{1}{1+(x-x_0{)}^2}$, where $x_0\notin X$ and $x_0$ is a limit point of $X$.
$X$ is not bounded, $f(X)$ does not contain its supremum/infimum: $f(x)=\frac{x^2}{1+x^2}$. ❏

Proof of 3
Since $X$ is not compact, we know $X$ has an infinite set $S$ with no limit point in $X$. But, since $X$ is bounded, it is contained in a closed interval $[-M,M]$, and so is $S$. Since closed intervals in $\mathbb{R}$ are compact, $S$ must have a limit point $x_0\in [-M,M]$. It follows that $x_0$ is also limit point of $X$. Now, define $f(x)=\frac{1}{x-x_0}$, and we are done. ❏