1
$\begingroup$

I think the answer should be yes, and I am trying to prove the positive answer.


Let $(X,T)$ be the topological space, and $G=\{ G_{\alpha} \}$ be our countable basis and $\mathbb{B}$ be the open ball basis.

to proceed, I think we should take balls and then cover with the countable basis.

For any $B in \mathbb{B}$, we can write

$$ B = \cup_{j} G_j$$

If the siye of $\mathbb{B}$ is uncountable, then we have covered uncountably many basis elements using countably many elements....which means...? I think there is some sort of repetition going, but I can't pin point it down.

I think somehow from the above we can somehow say that we can pull a countable number of balls which works from the above but it is not clear.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

We first show the following fact:

If a space $X$ has a countable basis $\{ G_n \}_{n \in \mathbb{Z}^+}$, then every basis $\{ B_\alpha \}_{\alpha \in J}$ contains a countable basis.

Consider each pair $i, j \in \mathbb{Z}^+$. If there exists some $\alpha \in J$ such that $G_i \subseteq B_\alpha \subseteq G_j$, then choose one such $\alpha$ arbitrarily and let $C_{i, j} = B_\alpha$. It is easy to verify that $\{ C_{i, j} \}$ is a basis:

  1. For any element $x \in X$, some element $G_{n_x}$ must contain it, since $\{ G_n \}$ is a basis. Since $\{ B_\alpha \}$ is a basis, there must be some element $B_{\alpha_x}$ such that $x \in B_{\alpha_x} \subseteq G_{n_x}$. Then, again since $\{ G_n \}$ is a basis, there must be some $G_{n'_x}$ such that $x \in G_{n'_x} \subseteq B_{\alpha_x} (\subseteq G_{n_x})$. Then letting $i_x = n'_x$ and $j_x = n_x$, the existence of $B_{\alpha_x}$ witnesses that $C_{i_x, j_x}$ exists and contains $x$.

  2. Let $C_{k, l}, C_{m, n}$ be arbitrary elements of the collection $\{ C_{i, j} \}$; let $z$ be in their intersection. Here, $U = C_{k, l} \cap C_{m, n}$ is a neighborhood of $z$. Proceed just as in the previous step, using the fact that $\{ G_n \}$ and $\{ B_\alpha \}$ are bases, to guarantee $C_{i_z, j_z} \in \{ C_{i, j} \}$ such that $z \in C_{i_z, j_z} \subseteq U$.

Obviously, the basis $\{ C_{i, j} \}_{i, j \in \mathbb{Z}^+}$ is countable.

Applying this result to your problem, consider the basis $\mathcal{B} = \{ B_d (x, \varepsilon) | x \in X, \varepsilon > 0 \}$ for the metric topology induced by $d$. It may be uncountable, but the existence of a countable basis $\{ G_n \}$ guarantees that $\mathcal{B}$ contains a countable basis (that consists entirely of ball neighborhoods).

$\endgroup$
2
  • 1
    $\begingroup$ Excellent work. $\endgroup$
    – Babu
    May 15 at 18:17
  • $\begingroup$ @trystwithfreedom Glad to help! $\endgroup$
    – K. Jiang
    May 15 at 20:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .