# Does the existence of a countable basis on a metric topology imply existence of a countable collection of balls which is also a basis?

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.

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).

• Excellent work.
– Babu
May 15 at 18:17
• @trystwithfreedom Glad to help! May 15 at 20:33