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Let $(X,d)$ be a metric space. We say that $X$ is separable if $X$ contains a countable dense subset.

We say that $X$ is second countable if $X$ has a countable base. The following fact is well-known.

Fact: If $X$ is a separable metric space, then any subset $Y$ of $X$ is also separable.

One main idea to prove above fact is that subset of a second countable space is again second countable.

Question: Let $X=\mathbb{R}$ and $d$ be euclidean norm(that is, absolute value $|\cdot|$). Is $\mathbb{R}\setminus\mathbb{Q}$ separable?

If we use the above fact, then $\mathbb{R}\setminus\mathbb{Q}$ should be separable as $\mathbb{R}$ contains $\mathbb{Q},$ which is a countable dense subset of $\mathbb{R},$ and $\mathbb{R}\setminus\mathbb{Q}$ is a subset of $\mathbb{R}.$

However, I do not see immediately a countable dense subset of $\mathbb{R}\setminus \mathbb{Q},$ as $\mathbb{R}\setminus\mathbb{Q}$ is uncountable, though it is dense in $\mathbb{R}.$

Any hint would be appreciated.

Community
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Idonknow
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3 Answers3

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You can construct a set based on the rationals which is exclusively contained in $\mathbb{R} \setminus \mathbb{Q}$.

Kurtland Chua
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Select any countable basis $\mathscr B$ for $\mathbb R$ (e.g., balls with rational centers and rational radii).

Select a point $x_B\in B\setminus\mathbb Q$ for each $B\in\mathscr B$.

Then $\{x_B:B\in\mathscr B\}$ is a countable dense subset of $\mathbb R\setminus \mathbb Q$.

MPW
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Let $r \in \mathbb R \setminus \mathbb Q$.

  • Let $ l \in \mathbb R$. Note that there exist rationals $q_n \to l - r$. However, this implies $q_n + r\to l$.

  • The above implies that the set $\mathbb Q + r = \{q + r : q \in \mathbb Q\}$ is dense in $\mathbb R$.

  • However, since $r$ is irrational, we have that $\mathbb Q + r \subset \mathbb R \setminus \mathbb Q$.

  • We have already shown that every element of $\mathbb R$ is approximable by elements of $\mathbb Q + r$. Since $\mathbb R \setminus \mathbb Q$ is just a subset of $\mathbb R$, the same property holds with $\mathbb R \setminus \mathbb Q$ as well.

  • Since $\mathbb Q +r$ is countable and dense in $\Bbb R \setminus \Bbb Q$, the separability of $\mathbb R \setminus \mathbb Q$ follows.

Sarvesh Ravichandran Iyer
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