This is a question from Rudin's Principles. Chapter 2, question 22.
The question reads: "A metric space is called $separable$ if it contains a countable dense subset. Show that $\mathbb{R}^k$ is separable. Hint: Consider the set of points which have only rational coordinates."
The answer starts with: "We need to show that every non-empty open subset $E$ of $\mathbb{R}^k$ contains a point with all coordinates rational." and then does just that, but I'm not sure how that addresses what the question is asking.
I know the definition of a dense subset. For a metric space $X$ and $E\subset{X}$, $E$ is dense in $X$ if every point of $X$ is a point of $E$ or a limit point of $E$ (or both).
And I know that saying a set is countable means that the set has the same cardinality as the natural numbers or in other words could be put into one-to-one correspondence with the naturals.
Combining the two definitions to get definition of a countable dense subset is pretty straightforward.
And I understand that the rationals are countable. I also understand that the rationals are dense in $\mathbb{R}$ which implies that $\mathbb{Q}^k$ is dense in $\mathbb{R}^k$.
But I don't know how showing that "every non-empty open subset $E$ of $\mathbb{R}^k$ contains a point with all coordinates rational" proves that $\mathbb{R}^k$ has a countable dense subset.
What am I missing here?
Is there another way to prove this?
Best Answer
It is a general result in topology that given a finite collection of topological spaces $\{A_i\}_{i=1}^n$, if $A=\prod_{i=1}^nA_i$ then
$$\overline{A}=\prod_{i=1}^n\overline{A_i}\;.$$
We already know that $\overline {\mathbb Q} = \mathbb R$, so $\overline {\mathbb Q^k} = \mathbb R^k$. Hence $\mathbb R^k$ is separable.