CW-complex, Hausdorff space, second-countable space, sober space
connected space, locally connected space, contractible space, locally contractible space
A topological space is separable if it has a countable dense subset.
To be explicit, $X$ is separable if there exists an infinite sequence $a\colon \mathbb{N} \to X$ such that, given any point $b$ in $X$ and any neighbourhood $U$ of $b$, we have $a_i \in U$ for some $i$.
A second-countable space is separable and first-countable, but the converse need not (see Steen Seebach Example 51 ).
Many results in analysis are easiest for separable spaces. This is particularly true if one wishes to avoid using strong forms of the axiom of choice or to be predicative over the natural numbers.
A metric space $X$ is separable iff every open set is a countable union of balls.
One direction is easy: if $x_i$ is a countable dense subset of $X$ and $r_j$ is an enumeration of the rationals, then the balls $B_{r_j}(x_i)$ is a countable base ($X$ is a second-countable space), and in particular every open set is a union of a family of such balls that is at most countable.
The other direction is trickier. (This is based on a MathOverflow discussion, which for the moment we record with little adaptation.) Suppose $X$ is not separable; construct by recursion a sequence $x_\beta$ of length $\omega_1$ such that no $x_\beta$ lies in the closure of the set of its predecessors $x_\alpha$ (each such set is countable and therefore not dense, so such $x_\beta$ outside its closure can be found at each stage).
Therefore, for each $x_\beta$ we may choose a rational radius $r_\beta$ such that the ball $B_{r_\beta}(x_\beta)$ contains no predecessor $x_\alpha$. There are uncountably many $\beta$, so some rational radius $r$ was used uncountably many times. The collection of $x_\beta$ for which $r_\beta = r$ forms another $\omega_1$-sequence which we now use instead of the first; since all the radii are the same, this sequence has the property that each $B_r(x_\beta)$ contains no other $x_\alpha$, no matter whether $\alpha$ appears before or after $\beta$ in the sequence.
Provided that there uncountably many such $x_\alpha$ that are non-isolated in $X$, we can easily construct an open set $U$ that is not a countable union of balls. For around each such $x_\alpha$ we can find a small ball $B_{s_\alpha}(x_\alpha)$ of radius at most $r/2$, such that $B_r(x_\alpha) \backslash B_{s_\alpha}(x_\alpha)$ is inhabited by a point $p_\alpha$. Put $U = \bigcup_\alpha B_{s_\alpha}(x_\alpha)$. No ball $B_s(x_\alpha)$ interior to $U$ can contain another $x_\beta$, since for that we would need $s \gt r$ and such $B_s(x_\alpha)$ would then include $p_\alpha$, which is excluded from $U$ by design. Thus, since there are uncountably many $x_\beta$, we need uncountably many balls to cover $U$.
We may therefore assume that there are at most countably many non-isolated $x_\beta$. Discard these, so without loss of generality we may suppose all the $x_\beta$ are isolated points of $X$ and that for some fixed $r$ the ball $B_r(x_\beta)$ contains no other $x_\alpha$. For each $\beta$, let $t_\beta$ be the supremum over all $t$ such that $B_t(x_\beta)$ is countable. It follows that $B_{t_\beta}(x_\beta)$ is itself countable, as is $\{\alpha: \alpha \lt \beta\}$. At each stage $\beta$, there is a countable set $C_\beta$ disjoint from $B_{t_\beta}(x_\beta) \cup \{\alpha: \alpha \lt \beta\}$ such that for each $s \gt t_\beta$, there exists $y \in C_\beta$ with $d(x_\beta, y) \lt s$. By transfinite induction, we can construct a cofinal subset $I$ of $\omega_1$ such that $x_\beta \notin C_\alpha$ whenever $\alpha \lt \beta$. The set $Y = \{x_\alpha: \alpha \in I\}$ is open in $X$, and we claim that it is not a countable union of balls. For suppose otherwise. Let $F$ be such a countable family of balls; then there is some minimal $\alpha$ for which $B_t(x_\alpha) \in F$ is uncountable, so that $t \gt t_\alpha$. By construction of $C_\alpha$, there exists $y \in C_\alpha \cap B_t(x_\alpha)$. But since $Y = \bigcup F$, we have that $y = x_\beta$ for some $\beta \in I$, and this contradicts our condition on $I$.
separable space