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I have a topology question regarding the proof that $\mathbb{R}\setminus \mathbb{Q}$ is not an $F_\sigma$.

The proof is very informal and would like to receive some formal explanation because I could not follow it at all.

So let us assume $\mathbb{R} \setminus \mathbb{Q} = \bigcup_{i}^{\infty}F_i$ such that $F_i$ is closed in $\mathbb{R}$ for every $i$. $\mathbb{Q} = \bigcup_{k}^{\infty}\{q_k\}$ Now $\mathbb{R} = (\bigcup_{i}^{\infty}F_i) \cup (\bigcup_{k}^{\infty}\{q_k\})$.

The supposed proof comes from the fact that one of these closed sets needs to contain an open set.

I can't see why one of these sets would have to contain an open set from everything written above.

Clarification would be greatly appreciated, or perhaps a more formal approach (I was not able to find one after browsing the web for a while).

Thanks!

Martin Sleziak
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Joshua
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    Baire category theorem. – Andrés E. Caicedo Dec 30 '13 at 23:20
  • HINT: Search the site. It has been covered countlessly many times (either directly, or by proving that $\Bbb Q$ is not $G_\delta$). – Asaf Karagila Dec 30 '13 at 23:24
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    This is listed among related questions (as the first one): http://math.stackexchange.com/questions/26311/proof-that-mathbbr-setminus-mathbbq-is-not-an-f-sigma-set See also: http://math.stackexchange.com/questions/69451/how-to-show-that-mathbbq-is-not-g-delta – Martin Sleziak Dec 31 '13 at 10:47
  • The proof you outline uses the Baire Category Theorem. I do not know offhand any proof of this fact not using the BCT (or containing essentially a re-proof of the BCT). – GEdgar Dec 31 '13 at 16:25

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