0.1  (§20) The metric topology

• Def. $(X,𝒯)$ is metrizable if $𝒯={𝒯}_d$ for some metric $d$.

• Def. $(X,d)$ bounded if $d\le M$ throughout; $\mathrm{diam}(X)=sup\{d(x,y)\}$.

• $\mathrm{★}$ Thm. $\overline{d}(x,y)=\min \{d(x,y),1\}$ is a bounded metric inducing the same topology as $d$.

  Proof. Triangle inequality by cases. Same topology because $\epsilon$-balls with already form a basis, and there $d$ and $\overline{d}$ agree. Moral: boundedness is metric data, not topological data.

• Def. On ${ℝ}^n$:  $\Vert x\Vert =\sqrt{x_1^2+\mathrm{\cdots }+x_n^2}$, $d(x,y)=\Vert y-x\Vert$ (Euclidean); ${\Vert x\Vert }_{\mathrm{\infty }}={\max }_i|x_i|$, $\rho (x,y)={\Vert y-x\Vert }_{\mathrm{\infty }}$ (square metric).

• Thm. $d$ and $\rho$ induce the same topology on ${ℝ}^n$.

• Def. ${ℝ}^J=\{x:J\to ℝ\}$;  ${ℝ}^{\omega }$ for $J=\mathbb{N}$.

  Note. Neither $\Vert x\Vert$ nor ${\Vert x\Vert }_{\mathrm{\infty }}$ is defined for all of ${ℝ}^{\omega }$ — hence the need to truncate with $\overline{d}$.

• Def. Uniform metric on $X^J$, $(X,d)$ metric: $\overline{\rho }(x,y)=sup\{\overline{d}(x_{\alpha },y_{\alpha })\mid \alpha \in J\}.$ The induced topology is the uniform topology.

• Thm. On ${ℝ}^J$: uniform topology is finer than the product topology (strictly, for $J$ infinite).

  Proof. Given a basis box $B$ constrained at ${\alpha }_1,\mathrm{\dots },{\alpha }_n$, take $\epsilon ={\min }_i{\epsilon }_i$; then $B_{\overline{\rho }}(x,\epsilon )\subset B$.

• $\mathrm{★}$ Thm. The product topology on ${ℝ}^{\omega }$ is metrizable: $D(x,y)=sup\left\{\overline{d}(x_n,y_n)/n\right|n\in \mathbb{N}\}$ induces it.

  Note. The $1/n$ damping makes all but finitely many coordinates irrelevant at scale $\epsilon$ — exactly matching “$U_n\ne ℝ$ for finitely many $n$”.

0.2  (§21) The metric topology, continued

• Def. $X$ has a countable basis at $x$ if there are neighborhoods ${\{B_n\}}_{n=1}^{\mathrm{\infty }}$ of $x$ such that every neighborhood of $x$ contains some $B_n$. $X$ is first-countable if this holds at every point. May assume $B_1\supset B_2\supset \mathrm{\cdots }$ (replace $B_n$ by $B_1\cap \mathrm{\cdots }\cap B_n$).

• Lem. Every metric space is first-countable  — take $B_d(x,1/n)$.

• $\mathrm{★}$ Lem. (Sequence lemma.) If some sequence in $A$ converges to $x$, then $x\in \overline{A}$. If $X$ is metrizable, the converse holds.

  Proof. Converse: pick $x_n\in A\cap B_d(x,1/n)$.

• $\mathrm{★}$ Thm. $f$ continuous $\Rightarrow$ ($x_n\to x$ gives $f(x_n)\to f(x)$). If $X$ is metrizable, the converse holds.

  Proof. Converse: show $f(\overline{A})\subset \overline{f(A)}$ using the sequence lemma.

• $\mathrm{★}$ Lem. ${ℝ}^J$ with $J$ uncountable is not metrizable.

  Proof. Let $A=\{x\mid x_{\alpha }\ne 1\text{ for finitely many }\alpha \}$ and $x=0$. Then $0\in \overline{A}$, but no sequence in $A$ converges to $0$: the union of countably many finite “support” sets misses some $\beta \in J$, and ${\pi }_{\beta }^{-1}(-\frac{1}{2},\frac{1}{2})$ separates. So the sequence lemma fails.