Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Lecture 18: Connectedness, compactness and the Mean value theorem

Let $X,Y$ be topological spaces and let $f:X\to Y$ be a continuous function. Let $E\subseteq X\text{.}$

(a)	If $E$ is cover compact then $f\left(E\right)$ is cover compact.
(b)	If $E$ is connected then $f\left(E\right)$ is connected.

Let $f:X\to ℝ$ be continuous function where $X$ is a compact metric space. Then $f$ attains a maximum and minimum value, i.e. there exist $$a\in X\text{such that}f\left(a\right)=\text{inf}\left\{f\left(x\right)\hspace{0.17em}\right|\hspace{0.17em}x\in X\}$$ and $$b\in X\text{such that}f\left(b\right)=\text{sup}\left\{f\left(x\right)\hspace{0.17em}\right|\hspace{0.17em}x\in X\}\text{.}$$

Let $A\subseteq ℝ\text{.}$

(a)	$A$ is connected if and only if $A$ is an interval.
(b)	$A$ is compact and connected then $A$ is a closed bounded interval.

(Rolle's theorem) $f:[a,b]\to ℝ$ with $f\left(a\right)=f\left(b\right)\text{.}$

(Mean value theorem) $f:[a,b]\to ℝ\text{.}$ There exists $c\in [a,b]$ with $$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\text{.}$$

		Sketches of proofs.
	(a) cover compact $\Rightarrow$ sequentially compact To show: not sequentially compact $\Rightarrow$ not cover compact. Let $a_1,a_2,\dots$ be a sequence in $A$ with no cluster point. For each $a\in A$ let ${\epsilon }_a\in {ℝ}_{>0}$ be such that $B(a,{\epsilon }_a)\cap \{a_1,a_2,\dots \}$ is finite. Then $\left\{B(a,{\epsilon }_a)\hspace{0.17em}\right|\hspace{0.17em}a\in A\}$ is an open cover of $A$ with no finite subcover. So $A$ is not cover compact. (b) sequentially compact $\Rightarrow$ cover compact To show: not cover compact $\Rightarrow$ not sequentially compact. Let $𝒮$ be a cover of $A$ with no finite subcover. Let $a_1\in A$ and $S_1\in 𝒮$ with $a_1\in S_1\text{.}$ Let $a_2\in A\backslash S_1$ and $S_2\in 𝒮$ with $a_2\in S_2\text{.}$ Let $a_3\in A\backslash (S_1\cup S_2)$ and $S_3\in 𝒮$ with $a_3\in S_3\text{.}$ $⋮$ Then $a_1,a_2,a_3,\dots$ is a sequence in $A$ with no cluster point. (A cluster point $a$ would have an $S\in 𝒮$ with $a\in S,$ and so $S$ is a neighbourhood of $a$ and would contain all but a finite number of $a_i$ and so $S_1\cup S_2\cup \cdots \cup S_N\cup S$ would be a finite cover??) $\square$

Notes and References

These are a typed copy of Lecture 18 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 27, 2014.

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