Two questions on Munkres -Topology: product and subspace of separable spaces

I have two questions:

If $X$ is a countable product of spaces having countable dense subsets then does $X$ have a countable dense subset?

Let $X$ $=\prod_{i=1}^\infty X_i$. Let $D_i$ denote the countable dense subset of the space $X_i$. Then $\scr D$ $=\prod_{i=1}^\infty D_i$ is a countable set as Cartesian product of countable sets is countable. Secondly $\overline {\prod _{i=1}^\infty D_i}=\prod _{i=1}^\infty \overline {D_i}=X$. Hence $\scr D$ is the required set. Is this correct?

Is a subspace of a separable topological space separable?

$\Bbb R$ has $\Bbb Q$ as countable dense set, but $\Bbb Q^c$ does not have a countable dense subset. How should I prove this fact? Am I right?

Solution 1:

Both of your arguments are wrong, I’m afraid. The Cartesian product of countably infinitely many countably infinite sets is uncountable, so $\prod_kD_k$ is not in general a countable subset of $X$, though it is a dense subset. It is nonetheless true that $X$ is separable. For each $k\in\Bbb N$ fix a point $p_k\in D_k$, and let

$$D=\left\{\langle x_k:k\in\Bbb N\rangle\in\prod_kD_k:\exists m\in\Bbb N\,\forall k\ge m(x_k=p_k)\right\}\;;$$

now show that this set $D$ is both countable and dense in $X$.

For the second question, $\Bbb R$ has a countable base, so every subspace of $\Bbb R$ has a countable base and therefore a countable dense subset. In particular, $\Bbb R\setminus\Bbb Q$ is separable. (A specific example of a countable dense subset of the irrationals is $\{q+\sqrt2:q\in\Bbb Q\}$.

It is true, however, that a subset of a separable space need not be separable. In fact, you can build an example by changing the topology of $\Bbb R$. Let each $x\in\Bbb Q$ be isolated. For $x\in\Bbb R\setminus\Bbb Q$ and $n\in\Bbb Z^+$ let $B_n(x)=\left(x-\frac1n,x+\frac1n\right)$, and let $\{B_n(x):n\in\Bbb Z^+\}$ be a local base at $x$.