Lecture 7

$(X,d)$ ball topology Say $Y$ is a closed set in $X$ (in other words, $Y^{\complement }$ open).

Say we have a sequence $\{X_n\}$ in $Y$ and that $x_n\to x\in X$. Then $x\in Y$.

Proof

Let’s say $x\notin Y$. Which is a contradiction. Because $x$ cannot converge outside $Y$ QED.

$(Y,d)$ complete if $(X,d)$ is complete.

Proposition

Say $X$ is a topological space, that is Hausdorff.

If $B\subset X$ is compact, then $B$ is closed.

Proof

Hausdorff ⇒ $A_y\cap B_y=\varnothing$ , both are open. Then $\{B_y{\}}_{y\in B}$ is an open cover of $B$

Compact ⇒ Pick out finite subcover $\{B_{y_i}{\}}_{i=1}^n$ of $B$. Then $x\in \stackrel{\text{Open}}{\overbrace{{\cap }_{i=1}^n{A}_{y_i}}}\subset B^{\complement }$ since ${\cap }_{i=1}^n{A}_{y_i}\cap B_{y_j}=\varnothing$ (for any $j$)

So $B^{\complement }$ is open, in other words, $B$ is closed.

QED.

Heine-Borel Theorem

Say $A\subset {\mathbb{R}}^n$. Then $A$ is compact $⟺$ $A$ is closed and bounded.

Proof

For →: $A$ is closed since ${\mathbb{R}}^n$ is Hausdorff and $A$ is compact, see the previous result. It is bounded since we can cover it by finitely many balls.

For ←: Assume first that $A$ is an $n$-cube with boundary included. Say $A$ is not compact. If an open cover of $A$ has no finite subcover, then by halving sides of cubes we get a sequence of cubes contained in each other, each having no finite subcover. The centres of these cubes form a Cauchy sequence with a limit $x\in A$. $d(x_n,x_m)<\epsilon ,\forall n,m>N$. Show: ${\mathbb{R}}^n$ is complete. Any neighbourhood of $x$ from the cover will obviously contain a small enough cube, and will be a finite subcover. But this is a contradiction.

QED.

$A^{\complement }$ open. Say $\{{\cup }_n\}$ is an cover of $A$. Then $\{{\cup }_n\},A^{\complement }$ is an open cover of the $n$-cube. $\{{\cup }_{n_i}{\}}_{i=1}^N,A^{\complement }$ cover of the $n$-cube. Then $\{{\cup }_{n_i}{\}}_{i=1}^N$ will cover $A$.

Cor

${\mathbb{R}}^n$ is locally compact Hausdorff, and $\sigma$-compact

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Complex Vector Space Exercise

Show that finite dimensional the complex vector space $V⟹V\simeq {\mathbb{C}}^n$ for some $n$. $x,y\in V$ $⟹x+y\in V$ $a\times x\in V,a\in \mathbb{C}$ ${\mathbb{C}}^2$: $(x,y)+(z,w)\equiv (x+z,y+w)$ $a\times (x,y)\equiv (ax,ay)$ Linear map:

Proof

$n=\dim V$. Say $\{v_i{\}}_{i=1}^n$ is a linear basis for $V$. Define bijective linear map $L:{\mathbb{C}}^n\to V$ by $L((x_1,\dots ,x_n))={\Sigma }_{i=1}^n{x}_i\times v_i$. It is linear; $L(a(x_1,\dots ,x_n)+b(y_1,\dots ,y_n))=L((a{x}_1+b{y}_1,\dots ,a{x}_n+b{y}_n))={\Sigma }_{i=1}^n(a{x}_i+b{y}_i)v_i=$ ${\Sigma }_{i=1}^n(a{x}_i{v}_i+b{y}_i{v}_i)=a\times {\Sigma }_{i=1}^n{x}_i{v}_i+b\times {\Sigma }_{i=1}^n{y}_i{v}_i=$ $a\times L((x_1,\dots ,x_n))+b\times L((y_1,\dots ,y_n))$

Surs. $v\in V$. Basis $⟹\exists \underset{\in \mathbb{C}}{\underbrace{x_i}}$ such that $v={\Sigma }_{i=1}^n{x}_i{v}_i=L((x_1,\dots ,x_n))$. So $v\in \text{lm}L$

Injective Say $L((x_1,\dots ,x_n))=L((y_1,\dots ,y_n))$ then ${\Sigma }_{i=1}^n{x}_i{v}_i={\Sigma }_{i=1}^n{y}_i{v}_i\stackrel{basis}{\to }{x}_i=y_i,\forall i$, so $x=y$ Injective for linear map $L⟺\ker L=\{0\}\equiv \{x\in \mathbb{C}|L(x)=0\}$.

$\ker L$ is a vector space $\text{lm}L$ is a vector space

$L=0\to \ker L={\mathbb{C}}^n$ $0\in V$ (vectors spaces have to start from the origin, as that is where vectors themselves start from).

Metric Space Exercise

Consider the unit circle $\mathbb{T}=S^1=\{(x,y)\in {\mathbb{R}}^2|x^2+y^2=1\}$ $\mathbb{T}=\{z\in \mathbb{C}||z|=1\}$

Say there is a circle $z=(x,y)$, $x^2+y^2=1$ This is a metric space for $d(z,z^{\prime })$ = arclength between $z$ and $z^{\prime }$.

Check:

1. $d(z^{\prime },z)=d(z,z^{\prime })$
2. $d(z,z^{\prime })=0⟺z=z^{\prime }$
3. $d(z,z^{\prime \prime })\le d(z,z^{\prime })+d(z^{\prime },z^{\prime \prime })$ $d^{\prime }(z,z^{\prime })$ = length of straight line between $z$ and $z^{\prime }$ $=\sqrt{(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2}=\Vert z-z^{\prime }\Vert$