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Math4201 Topology I (Lecture 14)
Metric topology
Product topology and metric topology
If X and Y are metrizable spaces, then the product space X\times Y is metrizable.
If X is metrizable, then the subspace A\subset X equipped with subspace topology is metrizable.
Proof
Let d be a metric on X. Then define d' be the restriction of d to A:
d':A\times A\to \mathbb{R}+
d'(x,y)=d(x,y)
x,y\in A\subseteq X
d' is a metric on A. Since the metric topology on A associated to d' is the same as the subspace topology.
Note that for any x\in A and r>0
B_r^{d}(x)\cap A=B_r^{d'}(x)\tag{*}
A basis for metric topology on A is given by:
\mathcal{B}=\{B_r^{d'}(x)|x\in A,r>0,r\in \mathbb{R}\}
A basis for the subspace topology on A is given by:
\mathcal{B}'=\{B_r^{d}(x)\cap A|x\in A,r>0,r\in \mathbb{R}\}
Since (*) holds, \mathcal{B}\subseteq \mathcal{B}'.
This shows that subspace topology on A is finer than the metric topology on A.
We need to show that for any B_r^{d}(x) with x\in X and y\in B_r^{d}(x)\cap A, we have r'>0 such that y\in B_r^{d'}(y)\subseteq B_r^{d}(x)\cap A.
Use triangle inequality, we have r'=r-d(x,y)>0 such that y\in B_r^{d'}(y)\subseteq B_r^{d}(x)\cap A.
Proposition on sequence and closure
Let X be a topological space and A\subseteq X. Then the following holds:
If there is a sequence \{x_n\}_{n=1}^\infty such that x_n\in A and x_n\to x, then x\in \overline{A}. (x may not be in A)
The reverse holds if X is a metric space. That is, if X is a metric space and x\in \overline{A}, then there is a sequence \{x_n\}_{n=1}^\infty such that x_n\in A and x_n\to x.
Example of non-metrizable space
For the second part of the claim
Let X=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots with the product topology over infinite product.
X=\text{Map}(\mathbb{N},\mathbb{R})=\{(x_1,x_2,x_3,\cdots)|x_i\in \mathbb{R},i\in \mathbb{N}\}.
The box topology on X is the topology generated by:
\mathcal{B}=\{(a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)\times \cdots|a_i,b_i\in \mathbb{R},i\in \mathbb{N}\}
It is easy to check that this is a basis.
Take A=\mathbb{R}^{> 0}\times \mathbb{R}^{> 0}\times \mathbb{R}^{> 0}\times \cdots=\{(x_1,x_2,x_3,\cdots)|x_i> 0\}.
\overline{A}=\mathbb{R}^{\geq 0}\times \mathbb{R}^{\geq 0}\times \mathbb{R}^{\geq 0}\times \cdots=\{(x_1,x_2,x_3,\cdots)|x_i\geq 0\}.
In particular, (0,0,0,\cdots)\in \overline{A}
Take a basis element B=(a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)\times \cdots\in \mathcal{B} containing (0,0,0,\cdots). This means that a_i<0<b_i Then (\frac {b_1}{2},\frac {b_2}{2},\frac {b_3}{2},\cdots)\in B\cap A.
This shows that (0,0,0,\cdots)\in \overline{A}.
We claim that there is no sequence \{x_n\}_{n=1}^\infty such that x_n\in A and x_n\to (0,0,0,\cdots).
We proceed by contradiction and suppose that there is such a sequence \{x_n\}_{n=1}^\infty such that x_n\in A and x_n\to (0,0,0,\cdots).
Since x_n\in A, we have v_n=(a_1^n,a_2^n,a_3^n,\cdots) with a_i^n>0 for all i\in \mathbb{N}.
Consider the following open set around (0,0,0,\cdots):
C=(-\frac{a_1^1}{2},\frac{a_1^1}{2})\times (-\frac{a_2^1}{2},\frac{a_2^1}{2})\times (-\frac{a_3^1}{2},\frac{a_3^1}{2})\times \cdots
We claim that v_n\notin C. Otherwise, we should have -\frac{a_i^j}{2}<a_j^n<\frac{a_j^j}{2} for all i\in \mathbb{N}.
But this doesn't hold for j=n. This shows that v_n\cancel{\to} (0,0,0,\cdots). Which is a contradiction.
For the first part of the claim. Let \{x_n\}_{n=1}^\infty\subset A converge to x. Then for any open set U of x, we have N such that x_n\in U for n\geq N.
In particular, U\cap A is non-empty because it has x_n for large enough n.
So x\in \overline{A}.