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?

  • 1
    $\begingroup$ $\mathbb Q^k$ is countable. If every open neighborhood of a point in $\mathbb R^k$ contains a point of $\mathbb Q^k$ then that point of $\mathbb R^k$ is a limit point of $\mathbb Q^k$. $\endgroup$ – John Douma Nov 15 '15 at 4:29
  • $\begingroup$ I guess you're missing the definition of "dense". $\endgroup$ – BrianO Nov 15 '15 at 5:33
  • $\begingroup$ It still requires a small proof to see that $\mathbb{Q}$ dense in $\mathbb{R}$ implies that $\mathbb{Q}^k$ is dense in $\mathbb{R}^k$. But then you are done: you have shown a countable dense subset, namely $\mathbb{Q}^k$, so by definition $\mathbb{R}^k$ is separable. $\endgroup$ – Henno Brandsma Nov 15 '15 at 10:46

It is a general result in topology that given a collection of topological spaces $\{A_i\}_{i=1}^n$, then if $A=\prod_{i=1}^nA_i$


We already know that $\overline {\mathbb Q} = \mathbb R$, so $\overline {\mathbb Q^k} = \mathbb R^k$. Hence $\mathbb R^k$ is separable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.