Lecture 11

Last lecture talked about Nets

Proposition

A topological space is Hausdorff $⟺$ each net converges to at most one point.

Proof

$\Rightarrow$:

“Easy”

$\Leftarrow$:

say $x\ne y$, therefore cannot be separated by disjoint neighbourhoods.

By the axiom of choice, pick $x_{(A,B)}\in A\cap B$, where $A$ and $B$ are neighbourhoods of $x$ and $y$ respectively. Consider the index set of pairs $(A,B)$ with $(A,B)\ge (A^{\prime },B^{\prime })$ if $A\subset A^{\prime }\cap B\subset B^{\prime }$.

This is a “ufos”, and $\{x_{(A,B)}\to x;x_{(A,B)}\to y$ which is a contradiction.

$\underset{\in A\cap B}{\underbrace{x_{(A,B)}}}\in A^{\prime }\forall \underset{⟹A\subset A^{\prime }\cap B\subset B^{\prime }}{\underbrace{(A,B)\ge (A^{\prime },B^{\prime })}}$

Proposition - Convergence in Topological Space

$f:X\to Y$ topological spaces with $x\in X$.

Then:
$f$ is continuous at $x⟺f(x_i)\to f(x)\forall x_i\to x$.
($\forall$ neighbourhood $B$ of $f(x)$ $\exists$ neighbourhood $A$ of $x$ such that $f(A)\stackrel{A\subset f^{-1}(B)}{\overbrace{\subset }}B$ )

Proof

$\Rightarrow$:

Suppose $x_i\to x$ and that $B$ is a neighbourhood of $f(x)$. Then there $\exists$ a neighbourhood $A$ of $x$ such that $f(A)\subset B$. Then $\{x_i\}$ (net) will eventually be in $A$. Then $\{f(x_i)\}$ will eventually be in $B$, so that means $f(x_i)\to f(x)$.

$\Leftarrow$:

(Going for a proof by contradiction)

Say $B$ is a neighbourhood of $f(x)$ such that every neighbourhood $A$ of $x$ intersects $X\setminus f^{-1}(B)$. Then $x$ belongs to the closure of $X\setminus f^{-1}(B)$. By previous proposition there $\exists$ net $\{x_i\}\subset X\setminus f^{-1}(B)$ such that $x_i\to x$.

$\text{to }x\text{circle}⟺x\in \overline{A}⟺x\leftarrow x_i\in A$

Definition

A subnet of a net $f:I\to X$ is a net $g:J\to X$ and a map $h:J\to I$ such that $g=f\circ h$ and such that $\forall i\in I$ $\exists j\in J$ with $h(j^{\prime })\ge i\forall j^{\prime }\ge j$.

Theorem

A space is compact $⟺$ every net has a converging subnet.

Examples

Example: Illustrates $\Rightarrow$

$x_n=n$

$x_n=\frac{1}{n}\in ⟨0,1]$

Example: Illustrates $\Rightarrow$

Consider the compact space $[0,1]$, say $\{y_n\subset [0,1,]\}$

Construct subsequence:
$x_1=y_1$ $x_2$ any $y_j$ in the half with infinitely many $y_i$s.
Pick $x_3$ to be in the half with infinitely many $y_i$s.

Get subsequence $\{x_n\}$ contained in more and more narrow intervals. So $\{x_n\}$ will be cauchy incomplete $[0,1]$, so it converges.

$\underset{\text{closed}⟹complete}{\underbrace{[0,1]}}\subset \underset{\to \text{complete}}{\underbrace{\mathbb{R}}}$

Exercises

Note

Question numbering is probably not the same as the ones from the exercises on Canvas. The numbering was done in order they appeared in the lecture.

Question 1

$f,g\text{contiuous}⟹f\circ g\text{continuous}$
$X\stackrel{g}{\to }Y\stackrel{f}{\to }Z$ and $X\stackrel{(f\circ g)(x)=f(g(x))}{\to }Z$

$f^{-1}(A)$ open in $Y$
for $A$ open in $Z$.

$g^{-1}(B)$ open in $X$
for $B$ open in $Y$

Is $(f\circ g{)}^{-1}(A)$ open in $X$ when $A$ is open in $Z$?

$(f\circ g{)}^{-1}(A)=g^{-1}(\underset{B}{\underbrace{f^{-1}(A)}})$

Take $A$ open in $Z$. Then $(f\circ g{)}^{-1}(A)=g^{-1}(f^{-1}(A))\stackrel{\text{claim}}{\leftarrow }\text{true for any subset of }A\text{ of }Z$ is open in $X$ since $B=f^{-1}(A)$ is open in $Y$, as $f$ is continuous.

But then $g^{-1}(B)$ is open in $X$ as $g$ is continuous. Hence $(f\circ g{)}^{-1}(A)$ is open in $X$, so $f\circ g$ is continuous.

Claim:

$(f\circ g{)}^{-1}(A)=g^{-1}(f^{-1}(A)),\forall A\subset Z$.

Proof:

Say $x\in (f\circ g{)}^{-1}(A)$, so $(f\circ g)(x)\in A$, or $f(g(x))\in A$, so $g(x)\in f^{-1}(A)$, so $x\in g^{-1}(f^{-1}(A))$.
If $x\in g^{-1}(f^{-1}(A))$, then $g(x)\in f^{-1}(A)$, so $f(g(x))\in A=(f\circ g)(x)$, so $x\in (f\circ g{)}^{-1}(A)$.

Alternatively:

Say we have $x_i\to x$ in $X$. Then $(f\circ g)(x)\stackrel{\text{Definition of }\circ }{\overbrace{=}}f(g(x))\stackrel{x_i\to x}{\overbrace{=}}f(g({\lim }_{i\to \infty }{x}_i))$ (Note: I’m assuming the $\lim$ is $i\to \infty$, as it was not defined in the lecture) $=f({\lim }_{i\to \infty }g(x_i))={\lim }_{i\to \infty }f(g(x_i))={\lim }_{i\to \infty }(f\circ g)(x_i)$.

Question 2

$X\simeq Y$

$\stackrel{\text{Compact}}{\overbrace{S}}\not\simeq \stackrel{\text{Not compact}}{\overbrace{\mathbb{R}}}$
${\mathbb{R}}^2$
$\mathbb{R}\to {\mathbb{R}}^2$
$x\mapsto (x,x)$

$\underset{\underset{\tan }{\underbrace{\simeq }}\underset{\simeq ⟨-\frac{\pi }{2},\frac{\pi }{2}⟩}{\underbrace{⟨0,1⟩}}}{\underbrace{\mathbb{R}}}\to S\subset {\mathbb{R}}^2$

$⟨-\frac{\pi }{2},\frac{\pi }{2}⟩\stackrel{\tan }{\to }\mathbb{R}=⟨-\infty ,\infty ⟩$
$S\not\simeq {\mathbb{R}}^2$ by compactness argument.
${\mathbb{R}}^2$ is ‘more connected’ than $\mathbb{R}$.

Question 3

Why does there not exist a continuous injection for $S\to \mathbb{R}$?

What happens then if you take the image $f:S\to \mathbb{R}$?
$f(S)\subset \mathbb{R}$ is both compact and connected, so $f(S)=[a,b]$ for some $a,b\in \mathbb{R}$.

Then $f:S\to [a,b]$ is continuous and bijective (as nothing is excluded from the image).

Since both spaces (Note: not sure if it is the correct link) are compact and Hausdorff, then $f^{-1}$ is also continuous, so $S\simeq [a,b]$. But removing one point leaves $S$ connected, but not $[a,b]$. So this is a contradiction.

Question 4

Show that the connected components of $\mathbb{Q}\subset \mathbb{R}$ consists of single points, and that one of these are open.

Topology on $\mathbb{Q}=\{\mathbb{Q}\cap A\mid A\text{ open in }\mathbb{R}\}$

$A\cap Q$

$\{p\},p\in \mathbb{Q}$
$\mathbb{Q}\cap ⟨-\epsilon +p,\epsilon +p⟩$ contains more points than $p$

$A\cap Q$

$\{p\},p\in \mathbb{Q}$
$\mathbb{Q}\cap ⟨-\epsilon +p,\epsilon +p⟩$ contains more points than $p$

In the graph above: $a\in \mathbb{R}\setminus \mathbb{Q}$.