Lecture 8

Defines in topological spaces:

• Connected
• Connected Component

Proposition

$[0,1]$ is connected.

Proof

$[0,1]=A\cup B$
$A\cap B=\varnothing$

Suppose open sets $A$ and $B$ disconnected it, and say $1\in B$.

Then every neighbourhood of $a\equiv \sup A\in [0,1]$ will intersect both $A$ and $B$.

Which is a contradiction as this is impossible!

QED.

2.2 Continuity

Recall: $f:\mathbb{R}\to \mathbb{R}$ is continuous at $x$ if ${\lim }_{y\to x}f(y)=f(x)$, and it would be discontinuous if ${\lim }_{y\to x}f(y)\ne f(x)$. More precisely, if $\forall \epsilon >0\exists \delta >0$ such that $|x-y|<\delta ⟹|f(x)-f(y)<\epsilon$ or $f(B_{\delta }(x))\subset B_{\epsilon }(f(x))$, or $B_{\delta }(x)\subset f^{-1}(B_{\epsilon }(f(x)))=\{y\in \mathbb{R}|f(y)\in B_{\epsilon }(f(x))\}$.

Definition

The definition is defined at: Continuous.

Proposition

A continuous $f:X\to \mathbb{R}$ with $X$ compact, it will attain both a maximum and a minimum.

Proof

Note: the continuous image of a compact set is compact since inverse images of an open cover will again be an open cover. Then the final step is to use the Heine–Borel theorem since $f(x)$ is compact in $\mathbb{R}$.

QED.

Something else

We say $f$ is continuous (at every $x$) if $f^{-1}(A)$ is open for every open $A\subset Y$.

Continuous images of connected spaces are connected. Again because inverse images of open subsets disconnecting an image would disconnect the domain.

So homeomorphisms preserve compactness and connectedness. They are topological invariants.