# Two questions on Munkres -Topology: praoduct 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?

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 $x$.

• Verify that this yields a topology on $\Bbb R$.
• Show that $\Bbb R$ is separable in this topology.
• Show that $\Bbb R\setminus\Bbb Q$ is not separable in this topology.
• Sir two questions 1. I am really not getting how $D$ has been constructed or how does it's elements look like ;Is it possible to elaborate a bit 2.For specifying a topology on a set we need to give the collection of open sets but you did not do that .Can you please do that ?I am not getting the concept of local base here. – Learnmore Feb 6 '16 at 16:55
• @learnmore: Let $p=\langle p_k:k\in\Bbb N\rangle$; this is a point in $\prod_kD_k$. $D$ consists of all of the points in $\prod_kD_k$ that differ from $p$ on at most finitely many coordinates. If each $X_k$ were $\Bbb R$, for instance, each $D_k$ were $\Bbb Q$, and each $p_k=0$, $D$ would be the set of rational sequences with only finitely many non-zero coordinates. \\ A local base at a point is also called a neighborhood base at the point; you may know it under this name. It’s a fundamental concept that I’d expect to find in any topology text. At any rate, what I’ve defined is a ... – Brian M. Scott Feb 6 '16 at 17:32
• ... genuine system of local bases, and it generates a topology $\tau$ in the following way: a set $U\subseteq\Bbb R$ is in $\tau$ if and only if for each $x\in U$ there is a $B\in\mathscr{B}_x$ such that $B\subseteq U$, where $\mathscr{B}_x=\{B_n(x):n\in\Bbb Z^+\}$ if $x$ is irrational, and $\mathscr{B}_x=\big\{\{x\}\big\}$ if $x$ is rational. – Brian M. Scott Feb 6 '16 at 17:34
• My attempt: As constructed $D$ becomes the finite product of countably many dense sets and hence countable. Take any basic open set may be $\scr U=$ $U_1\times U_2\times... \times U_n\times X\times ...X$ Now as each $U_i$ intersects $D_i$ we have $x_i\in U_i\cap D_i$ for all $1\le i\le n$ then $(x_1,x_2,...,x_n,p,p...,p)\in \scr U\cap D$ – Learnmore Feb 7 '16 at 2:57
• Is the arguement correct? – Learnmore Feb 7 '16 at 2:58