Every separable metric space has a countable basis

Reference: Topology by Hocking and Young, Dover Publications Inc., NY. (available in Amazon India also)

….As an example of the topological power of a metric, we give the following result. First, a set X of points in a space S is said to be dense in S if every point of S is a point or a limit point of X, that is, if $S = \overline{X}$ . A space is separable if it has a countable dense subset. For instance, $E^{n}$ , the Euclidean n-space is separable since the set of all points whose coordinates are all rational is countable and dense.

Theorem 1-5: Every separable metric space has a countable basis.

Proof of Theorem 1-5: Let M be a metric space with metric $d(x,y)$ and having a countable dense subset $X = \{ x_{i} \}$ . For each rational number $r>0$ and each integer $i>0$ , there is a spherical neighbourhood $S(x_{i},r)$ and the set $\mathcal{B}$ of all these is countable. We will show that $\mathcal{B}$ is a basis. (Recall definition of a basis: A subcollection $\mathcal{B}$ of open sets of a topological space S is a basis for S if and only if every open set in S is a union of elements of $\mathcal{B}$ ) (A countable basis for a space S is a basis that contains only countably many sets).Let p be any point of M and let O be an open set containing p. Then there is a positive number $\epsilon$ such that $S(p, \epsilon)$ is contained in O, by definition. There is a point $x_{i}$ of X such that $d(x_{i},p) < \epsilon/3$ since X is dense. Let r be a rational number satisfying $\epsilon/3<r<2\epsilon/3$ and consider $S(x_{i},r)$ . Certainly $S(x_{i},r)$ contains p and if y is any point of $S(x_{i},r)$ , then $d(y,p) \leq d(y,x_{i}) + d(x_{i},p) < \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$ .

Thus y is an element of $S(p, \epsilon)$ and so $S(x_{i},r)$ is an element of $\mathcal{B}$ that contains p and lies in O. It follows that O is a union of elements of $\mathcal{B}$ and that $\mathcal{B}$ is a basis for the topology of M.

QED.

Important remarks about the above theorem:

Without the assumption of metricity, Theorem 1-5 is not true. In $E^{n}$ , the Euclidean n-space, consider the set P of all points $(x,y)$ with $y \geq 0$ . Let a basis for P consist of (1) all interiors of circles of P but not touching the x-axis and (2) the union of a point on the x-axis and the interior of a circle tangent from above to the x-axis at that point. The set of points in P both of whose coordinates are rational is both countable and dense in P. But no element of the basis just defined contains two points on the x-axis. If there were a countable basis $\mathcal{B}$ for P, then each basis element above would be a union of elements of $\mathcal{B}$ . This would imply that there is a subcollection of $\mathcal{B}$ such that each point of the x-axis lies in one and only one element of that subcollection. This contradicts the fact the real numbers are uncountable. QED.

Cheers,

Nalin Pithwa

MScMathematics

Every separable metric space has a countable basis

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