1. ## countable first countable

True/False:
1. Is every countable first countable $\displaystyle T_3$-space metrizable?
2. Is every separable first countable $\displaystyle T_3$ space metrizable?

Attempt
1. countable + first countable $\displaystyle \Rightarrow$ second countable. second countable + $\displaystyle T_3$ $\displaystyle \Leftrightarrow T_4$. So by Urysohn Theorem, since this is second countable + $\displaystyle T_4$, this is metrizable. But this argument doesn't work b/c $\displaystyle T_3 \not \Rightarrow T_4$ in general. So is this false? What is the counterexample?
2. separable + first countable $\displaystyle \Rightarrow$ second countable. Now, the argument is the same as above.
Thanks.

2. Originally Posted by xianghu21
True/False:
1. Is every countable first countable $\displaystyle T_3$-space metrizable?
2. Is every separable first countable $\displaystyle T_3$ space metrizable?

Attempt
1. countable + first countable $\displaystyle \Rightarrow$ second countable. second countable + $\displaystyle T_3$ $\displaystyle \Leftrightarrow T_4$. So by Urysohn Theorem, since this is second countable + $\displaystyle T_4$, this is metrizable. But this argument doesn't work b/c $\displaystyle T_3 \not \Rightarrow T_4$ in general. So is this false? What is the counterexample?
2. separable + first countable $\displaystyle \Rightarrow$ second countable. Now, the argument is the same as above.
Thanks.
The Uryshon Metrization Theorem: Every second countable regular Hausdorff space is metrizable.

For (1), if the topological space X is countable, then a countable basis can cover X. Thus, X is second countable (If X is second countable, then it is automatically first countable). By Urysohn's metrization theorem, every second countable regular Hausdorff space is metrizable (I denoted $\displaystyle T_3$ as a regular Hausdorff space. Otherwise, you can also use the fact that every regular Lindelof space is normal. This issue is discussed here).
Thus, (1) is correct.

For (2), the counterexample can be a lower limit topology on R, denoted as $\displaystyle R_l$. $\displaystyle R_l$ has a countable local basis at each point $\displaystyle x \in R_l$, where $\displaystyle \{[x, x+1/n)\}_{n=1}^{\infty}$ is a countable local basis at x. It is separable since the rational numbers are dense in $\displaystyle R_l$.
$\displaystyle R_l$ has no countable basis. Let B be a basis for $\displaystyle R_l$. For each x in $\displaystyle R_l$, we choose a basis element $\displaystyle B_x$ of B such that $\displaystyle x \in B_x \subset [x, x+1)$. If $\displaystyle x \neq y$, then $\displaystyle B_x \neq B_y$, since $\displaystyle x=glb \text{ } B_x$ and $\displaystyle y=glb \text{ } B_y$. Thus, B must be uncountable. It follows that $\displaystyle R_l$ is not second countable.
Thus, (2) is not correct.

Some additional remarks: The regularity hypothesis in the Urysohn metrization theorem is a necessary one, but the second countable condition is not. You can use other conditions to check metrizability (link).