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 trivially it is first-countable). However, first-countability plus separability do not imply second-countability; a counterexample is $\mathbb{R}$ with the half-open topology (see Munkres, page 192), denoted $\mathbb{R}_l$.
An arbitrary product of separable spaces need not be separable, but a product of as many as a continuum number of separable spaces is separable (with the product computed in Top); a proof of the more general Hewitt-Marczewski-Pondiczery theorem can be found here. In particular, the space $\mathbb{R}^\mathbb{R}$ of all functions $\mathbb{R} \to \mathbb{R}$ under pointwise convergence is separable, but is not even first-countable (and thus not second-countable either). A first-countable space need not be separable; a simple example of that is a discrete space of uncountable cardinality.
Although an arbitrary product of separable spaces need not be separable, an arbitrary product of separable spaces does satisfy the countable chain condition (see there for more discussion). This is somewhat remarkable, since it is not necessarily true that even a finite product of spaces satisfying the countable chain condition also satisfies the countable chain condition (whether it does is independent of ZFC).
Subspaces of separable spaces need not be separable. Example: the product $\mathbb{R}_l \times \mathbb{R}_l$, also called the Sorgenfrey plane?, is separable, but the subspace defined by the equation $y = -x$ is uncountable and discrete and therefore not separable. However, open subspaces of separable spaces are separable.
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 use arguments predicative over the natural numbers. For example, the Hahn-Banach theorem for separable Banach spaces can be established using only mild forms of choice, e.g., dependent choice. More precisely, it can be established in the weak subsystem $WKL_0$ of second-order arithmetic; see Brown-Simpson.
A classical fact is
For a metric space $X$ the following are equivalent
The second property is implied by the first: given any dense subset $\{x_n \mid n\in \mathbb{N}\}$, form the countable system of sets $\{B_{1/m}(x_n) \mid n,m\in\mathbb{N}\}$. To see that this is indeed a base for the topology of $X$, take any open set $U\subset X$, a point $x\in U$ and a radius $1/k$ such that $B_{1/k}(x)\subset U$. By separability there is some $n$ such that $x_n\in B_{1/(2k)}(x)$ and therefore $x\in B_{1/(2k)}(x_i)\subset U$.
To show (2) implies (3), let $\{U_n\}$ be a countable base of the topology. Given any open cover $\{V_\lambda\}$ of $X$, we can form the index set $I\subset \mathbb{N}$ of those $n$ that are contained in some $V_\lambda$. By assumption $\bigcup_{i\in I} U_{i} = \bigcup_\lambda V_\lambda = X$. The axiom of countable choice provides now a section of $\bigsqcup_{i\in I} \{\lambda \mid U_i \subset V_\lambda\}\to I$.
Finally, we prove that (3) implies (1). Consider the open covers $\{B_{1/1}(x) \mid x\in X\}$, $\{B_{1/2}(x) \mid x\in X\}$, … From each extract a countable subcover corresponding to collection of centers $A_1, A_2, \ldots$. We claim that that the union $A_1\cup A_2\cup\ldots$ forms a dense set. Indeed, given any $y\in X$ and $n$ the point $x$ has to be contained in some $B_{1/n}(x)$ for some $x\in A_n$.
It is also possible to prove without the axiom of choice that (3) implies (2), analogous to the argument for (3)$\Rightarrow$(1).
Similar in spirit to (1)$\Leftrightarrow$(2) but less well-known is the following.
A metric space $X$ is separable iff every open set is a countable union of balls.
One direction is not hard: if $x_i$ is a countable dense subset of $X$ and $r_j$ is an enumeration of the rationals, then according to the proof of Theorem 1, the balls $B_{r_j}(x_i)$ form a countable base ($X$ is a second-countable space). Hence 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 construct an open set $U$ that is not a countable union of balls. For around each such $x_\alpha$ we can find a point $p_\alpha \neq x_\alpha$ within distance $r/4$ of $x_\alpha$; put $U = \bigcup_\alpha B_{d(x_\alpha, p_\alpha)}(x_\alpha)$. Notice that $p_\alpha \notin U$. If $B_s(x)$ is any ball contained in $U$, then $d(x, x_\gamma) \lt r/4$ for some $\gamma$. Supposing we have distinct $x_\alpha, x_\beta \in B_s(x)$, then $r \lt d(x_\alpha, x_\beta) \leq d(x_\alpha, x) + d(x, x_\beta) \lt s + s$, so $r/2 \lt s$. But then from $d(x_\gamma, p_\gamma) \lt r/4$ and $d(x, x_\gamma) \lt r/4$, we have $d(x, p_\gamma) \lt r/2 \lt s$ so that $p_\gamma \in B_s(x) \subset U$, a contradiction. We conclude that any ball contained in $U$ contains at most one $x_\alpha$, and so $U$ cannot be covered by countably many balls.
We are therefore left to deal with the case where 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 $x_\alpha$. Let $Z$ be this set of $x_\beta$. For each $\beta$, let $t_\beta$ be the supremum over all $t$ such that $B_t(x_\beta) \cap Z$ is countable. It follows that $B_{t_\beta}(x_\beta) \cap Z$ is itself countable, as is $\{\alpha: \alpha \lt \beta\}$. At each stage $\beta$, there is a countable set $C_\beta \subset Z$ disjoint from $B_{t_\beta}(x_\beta) \cup \{x_\alpha: \alpha \lt \beta\}$ 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 $\alpha, \beta \in I$ and $\alpha \lt \beta$. 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 $z \in C_\alpha \cap B_t(x_\alpha)$. But since $Y = \bigcup F$, we have that $z = x_\beta$ for some $\beta \in I$, and this contradicts our condition on $I$.
separable space