Math4201 Topology I (Lecture 12)

Let $X=\mathbb{R}$ and $d(x,y)=|x-y|$.

Check definition of metric space:

1. Positivity: $d(x,y)=|x-y|\geq 0$ and $d(x,y)=0$ if and only if $x=y$.
2. Symmetry: $d(x,y)=|x-y|=|y-x|=d(y,x)$.
3. Triangle inequality: $d(x,z)=|x-z|\leq |x-y|+|y-z|=d(x,y)+d(y,z)$ since $|a+b|\leq |a|+|b|$ for all $a,b\in \mathbb{R}$.

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Let $X$ be arbitrary. The trivial metric is $d(x,y)=\begin{cases} 0 & \text{if } x=y \\ 1 & \text{if } x\neq y \end{cases}$

Check definition of metric space:

1. Positivity: $d(x,y)=\begin{cases} 0 & \text{if } x=y \\ 1 & \text{if } x\neq y \end{cases}\geq 0$ and $d(x,y)=0$ if and only if $x=y$.
2. Symmetry: $d(x,y)=\begin{cases} 0 & \text{if } x=y \\ 1 & \text{if } x\neq y \end{cases}=d(y,x)$.
3. Triangle inequality use case by case analysis.

Let $X=\mathbb{R}$ and $d(x,y)=\begin{cases} 0 & \text{if } x=y \\ 1 & \text{if } x\neq y \end{cases}$

The balls of this metric space are:

$$B_r(x)=\begin{cases} \{x\} & \text{if } r<1 \\ X & \text{if } r\geq 1 \end{cases}$$

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Let $X=\mathbb{R}$ and $d(x,y)=|x-y|$.

The balls of this metric space are:

$$B_r(x)=\{(x-r,x+r)\}$$

This basis is the set of all open sets in $\mathbb{R}$, which generates the standard topology of $\mathbb{R}$.

Let’s check the two properties of basis:

1. $\forall x\in X$, $\exists B_r(x)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}$ such that $x\in B_r(x)$. (Trivial by definition of non-zero radius ball)
2. $\forall B_r(x),B_r(y)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}$, $\forall z\in B_r(x)\cap B_r(y)$, $\exists B_r(z)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}$ such that $z\in B_r(z)\subseteq B_r(x)\cap B_r(y)$.

Observe that for any $z\in B_r(x)$, then there exists $\delta>0$ such that $B_\delta(z)\subseteq B_r(x)$.

Let $\delta=r-d(x,z)$, then $B_\delta(z)\subseteq B_r(x)$ (by triangle inequality)

Similarly, there exists $\delta'>0$ such that $B_\delta'(z)\subseteq B_r(y)$.

Take $\lambda=min\{\delta,\delta'\}$, then $B_\lambda(z)\subseteq B_r(x)\cap B_r(y)$.

Trivial topology with at least two points is not Hausdorff, so it isn’t metrizable.

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Finite complement topology on infinite set is not Hausdorff.

Suppose there exists $x,y\in X$ such that $x\neq y$ and $x\in U\subseteq X$ and $y\in V\subseteq X$ such that $X-U$ and $X-V$ are finite.

Since $U\cap V=\emptyset$, we have $V\subseteq X-U$, which is finite. So $X-V$ is infinite. (contradiction that $X-V$ is finite)

So $X$ with finite complement topology is not Hausdorff, so it isn’t metrizable.

Let $x,y\in (X,d)$ and $x\neq y$. To show that $X$ is Hausdorff, it is suffices to show that there exists $r,r'>0$ such that $B_r(x)\cap B_r'(y)=\emptyset$.

Take $r=r'=\frac{1}{2}d(x,y)$, then $B_r(x)\cap B_r'(y)=\emptyset$. (by triangle inequality)

We prove this by contradiction.

Suppose $\exists z\in B_r(x)\cap B_r'(y)$, then $d(x,z)<r$ and $d(y,z)<r'$.

Then $d(x,y)\leq d(x,z)+d(z,y)<r+r'=\frac{1}{2}d(x,y)+\frac{1}{2}d(x,y)=d(x,y)$. (contradiction $d(x,y)<d(x,y)$)

Therefore, $X$ is Hausdorff.