Show that a metrizable space with a countable dense set has a countable basis.

My try:

Let $X$ be a metrizable space with a countable dense set $D$.

  • Consider for each $n\in \Bbb N;B(x,\dfrac{1}{n});x\in X$. Always we have that $B(x,\dfrac{1}{n})\cap D\neq \emptyset $. Let $a\in B(x,\dfrac{1}{n})\cap D$.
  • Claim $\scr B$=$\{B(a,\dfrac{1}{n});a\in D;n\in \Bbb N\}$ is a countable basis of $X$.
  • Proof:

Let $x$ be arbitrary and $U$ be an open set containing $x$. Then for some positive integer $k$ we have $B(x,\dfrac{1}{k})\subset U$. Then $B(x,\dfrac{1}{2k})\cap D\neq \emptyset $. Let $b\in B(x,\dfrac{1}{2k})\cap D\implies d(x,b)<\dfrac{1}{2k}$.

By Triangle inequality $x\in B(b,\dfrac{1}{2k})\subset B(x,\dfrac{1}{k})\subset U$

  • Is the proof correct? Please suggest required edits.
  • $\begingroup$ The first bullet is unnecessary (you don't use it anyway). Just state the claim under the second one. $\endgroup$ – Henno Brandsma Feb 6 '16 at 9:19

Actually you almost made it. In the end you proved that $x\in U$. This is true since you took $x$ and then $U$ to be an arbitrary open set containing $U$.

What you need to prove is that $$y\in B(b,\frac{1}{2k}) \qquad \Longrightarrow \qquad y\in B(x,\frac{1}{k}).$$ This you can prove by Triangle inequality.

  • $\begingroup$ It's indeed better to just prove separately that $B(b, \frac{1}{2k}) \subseteq B(x,\frac{1}{k})$ as you say. Then we have for $x \in U$ found a base element (namely $B(b,\frac{1}{2k})$) that contains $x$ and is a subset of $U$. This is what is required to show being a base. $\endgroup$ – Henno Brandsma Feb 6 '16 at 10:00
  • $\begingroup$ I never proved $x\in U$ .I proved what you have stated ;i did say that using $\Delta $ inequality $\endgroup$ – Learnmore Feb 6 '16 at 10:32

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