Proof that Metric Spaces are Second Countable? Metric spaces can be equipped with the topology given by the open sets (which in turn are defined with the help of open balls) such that metric spaces are topological spaces. Now, the definition of second countable we had in our lecture (where $(M, \tau)$ shall be a topological space):

$(M, \tau)$ is called second countable if its topology has a countable basis.

Now, Wikipedia has this inconspicuous sentence:

In second-countable spaces—as in metric spaces—compactness, sequential compactness, and countable compactness are all equivalent properties.

Question: So apparently, metric spaces are always second countable; however, how can we prove this? (For $M = \mathbb R^n$ it is clear, as the basis is given by open balls $B_r(p)$ with positive and rational radii $r$ and rational centres $p$, but I am not sure how the reasoning goes for arbitrary metric spaces.)
 A: No it's a matter of text parsing. The Wikipedia text could (or should) have stated: consider the following properties for a space $X$:

*

*$X$ is compact.

*$X$ is sequentially compact.

*$X$ is countably compact.

For general spaces these properties are not equivalent (see below too).
But if we know that $X$ is (as extra info) second countable, they are equivalent.
And also if we know $X$ is metrisable, they are equivalent as well.
So it points out that these properties become equivalent for two distinct subclasses of spaces, the second countable ones and the metrisable ones.
It does not mean to imply that these subclasses are otherwise related by implication.
And e.g. $X=\omega_1$ is countably compact and sequentially compact and not compact (and is thus neither second countable nor metrisable) and $X=[0,1]^{\Bbb R}$ is compact but not sequentially compact (and thus also neither second countable nor metrisable). For all spaces compact does imply countably compact by definition, of course.
A: That is not true; not every metric space is second countable.
What you have instead, as the linked Wikipedia says explicitly, is that "For metric spaces, however, the properties of being second-countable, separable ... are all equivalent."
See also Noah's comment for the language interpretation of your excerpt.
A: It is false. Take any uncountable set $X$ with the discrete metric $d(x,y) = 1$ for $x \ne y$, $d(x,x) = 0$.
