6.6 KiB
lecture, date
| lecture | date |
|---|---|
| 11 | 2025-02-13 |
Last lecture talked about Nets
Proposition
A Topological Space is Hausdorff \iff each Nets converges to at most one point.
Proof
\Rightarrow:
"Easy" !Drawing 2025-02-13 10.57.31.excalidraw
\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.
!Drawing 2025-02-13 11.02.27.excalidraw
\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 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) 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 \{ x_{i} \} \subset X \setminus f^{-1} (B) such that x_{i} \to x.
!Drawing 2025-02-13 11.26.08.excalidraw
\text{to }x \; \text{circle}\iff x \in \overline{A} \iff x \leftarrow x_{i} \in A
Definition
A Subnet of a Nets f: I \to X is a Nets 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 \iff every net has a converging subnet.
Examples
[!example] Example: Illustrates
\Rightarrowx_{n} = n!Drawing 2025-02-13 12.01.57.excalidraw
x_{n} = \frac{1}{n} \in \langle \, 0, 1 \, ]
[!example] Example: Illustrates
\RightarrowConsider the Compact space[ \, 0, \, 1 \,], say\{ y_{n} \subset [\, 0, \, 1, \, ] \}!Drawing 2025-02-13 12.07.00.excalidraw Construct subsequence:x_{1} = y_{1}x_{2}anyy_{j}in the half with infinitely many $y_{i}$s. Pickx_{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 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 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' 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 Topological Space (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 Sets.
Topology on \mathbb{Q} = \{ \mathbb{Q} \cap A \mid A \text{ open in } \mathbb{R} \}
!Drawing 2025-02-13 14.01.44.excalidraw
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
In the graph above: a \in \mathbb{R} \setminus \mathbb{Q}.