---
lecture: 11
date: 2025-02-13
---
Last lecture talked about [[Nets]]
# Proposition
A [[Topological Space|topological space]] is [[Hausdorff]] $\iff$ each [[Nets|net]] converges to at most one point.
## Proof
### $\Rightarrow$:
"Easy"
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### $\Leftarrow$:
say $x \neq y$, 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 $(A, B) \geq (A', B')$ if $A \subset A' \cap B \subset B'$.

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

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$$\underbrace{x_{(A,B)}}_{\in {A \cap B}} \in A' \forall \underbrace{(A, B) \geq (A', B')}_{\implies A \subset A' \cap B \subset B'}$$
# Proposition - Convergence in Topological Space
$f: X \to Y$ [[Topological Space|topological spaces]] with $x \in X$.

Then:
$f$ is continuous at $x \iff f(x_{i}) \to f(x) \; \forall x_{i} \to x$.
($\forall$ neighbourhood $B$ of $f(x)$ $\exists$ neighbourhood $A$ of $x$ such that $f(A) \overbrace{\subset}^{A \subset f^{-1}(B)} B$ )
## 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 $\{ x_{i} \}$ ([[Nets|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 $\exists$ [[Nets|net]] $\{ x_{i} \} \subset X \setminus f^{-1} (B)$ such that $x_{i} \to x$.

![[Drawing 2025-02-13 11.26.08.excalidraw.dark.svg]]
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$$\text{to }x \; \text{circle}\iff x \in \overline{A} \iff x \leftarrow x_{i} \in A$$
## Definition
A **[[Subnet|subnet]]** of a [[Nets|net]] $f: I \to X$ is a [[Nets|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 $h(j') \geq i \; \forall j' \geq j$.
## Theorem
A space is [[Compact|compact]] $\iff$ every net has a converging subnet.
### Examples
> [!example] Example: Illustrates $\Rightarrow$
> $$x_{n} = n$$
> ![[Drawing 2025-02-13 12.01.57.excalidraw.dark.svg]]
> $$x_{n} = \frac{1}{n} \in \langle \, 0, 1 \, ]$$

> [!example] Example: Illustrates $\Rightarrow$
> Consider the [[Compact|compact]] space $[ \, 0, \, 1 \,]$, say $\{ y_{n} \subset [\, 0, \, 1, \, ] \}$
> ![[Drawing 2025-02-13 12.07.00.excalidraw.dark.svg]]
> **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 Sequence|cauchy]] incomplete $[ \, 0, \, 1 \, ]$, so it converges.
> > [!note]
> > $$\underbrace{[\, 0, \, 1 \, ]}_{\text{closed} \implies complete} \subset \underbrace{\mathbb{R}}_{\to \, \text{complete}}$$

# 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
$f, \, g \; \text{contiuous} \implies f \circ g \; \text{continuous}$ 
$X \xrightarrow{g} Y \xrightarrow{f} Z$ and $X \xrightarrow{(f \circ g)(x) = f(g(x))} Z$

$f^{-1}(A)$ [[Open Sets|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 $Z$?

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

Take $A$ open in $Z$. Then $(f \circ g)^{-1}(A) = g^{-1}(f^{-1}(A)) \xleftarrow{\text{claim}} \text{true for any subset of } A \text{ of } Z$ 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 $(f \circ g)(x) \overbrace{=}^{\text{Definition of } \circ} f(g(x)) \overbrace{=}^{x_{i} \to x} f(g(\lim_{ i \to \infty }x_{i}))$ (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
$X \simeq Y$

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

$\underbrace{\mathbb{R}}_{\underbrace{\simeq}_{\tan} \underbrace{\langle \, 0, \, 1 \, \rangle}_{\simeq \langle \, -\frac{\pi}{2}, \, \frac{\pi}{2} \, \rangle}} \to S \subset \mathbb{R}^2$

$\langle \, -\frac{\pi}{2}, \, \frac{\pi}{2} \, \rangle \xrightarrow{\tan} \mathbb{R} = \langle \, - \infty, \, \infty \, \rangle$
$S \not\simeq \mathbb{R}^2$ by compactness argument.
$\mathbb{R}^2$ is 'more [[Connected|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|connected]], so $f(S) = [ \, a, \, b \, ]$ for some $a, \, b \in \mathbb{R}$.

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

Since both [[Topological Space|spaces]] (Note: not sure if it is the correct link) are [[Compact|compact]] and [[Hausdorff]], then $f^{-1}$ is also continuous, so $S \simeq [ \, a, \, b \, ]$. But removing one point leaves $S$ [[Connected|connected]], but not $[ \, a, \, b \, ]$. So this is a contradiction.
## Question 4
Show that the [[Connected|connected]] components of $\mathbb{Q} \subset \mathbb{R}$ consists of single points, and that one of these are [[Open Sets|open]].

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

![[Drawing 2025-02-13 14.01.44.excalidraw.dark.svg]]
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$A \cap Q$

$\{ p \}, \; p \in \mathbb{Q}$
$\mathbb{Q} \cap \langle - \varepsilon + p, \, \varepsilon + p \rangle$ contains more points than $p$

![[Drawing 2025-02-13 14.01.44.excalidraw.dark.svg]]
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$A \cap Q$

$\{ p \}, \; p \in \mathbb{Q}$
$\mathbb{Q} \cap \langle - \varepsilon + p, \, \varepsilon + p \rangle$ contains more points than $p$
![[Drawing 2025-02-13 14.06.25.excalidraw.dark.svg]]
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