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From some topology worksheets I found online that I am using to self-study: Let $I$ be the set of irrationals, and $d$ be the Euclidean metric. What is the category of $I$ in $(I, d)$? I'm sure it is of the second category, but I am unsure of how to prove it. At first I wanted to use the fact that the irrationals are of second category in the reals, but since I now don't have the fact that every neighborhood of some irrational contains a rational, I don't know to proceed.

Hanul Jeon
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Phillip Feldman
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  • It's more than that --- it's complement is actually first category. Note that it's complement is a countable union of nowhere dense sets (each singleton set is a nowhere dense set). – Dave L. Renfro Nov 12 '20 at 08:23
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    I just realized my previous comment is a bit overly simplistic, as I overlooked that you want the category of I in the space I, not the category of I in the space ${\mathbb R}.$ It's possible for a subset $E$ of $\mathbb R$ to be second category in the space $E$ and first category in the space $\mathbb R$ (even nowhere dense in ${\mathbb R}).$ For example, let $E$ be a finite subset of ${\mathbb R}.$ – Dave L. Renfro Nov 12 '20 at 10:53

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The space of all irrationals is completely metrizable, since is is a $G_\delta$ subset of $\mathbb{R}$. And so by the Baire category theorem it is of the second category.

freakish
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Lemma: The property of being a nowhere dense set is, so to say, "densely intrinsic", meaning that if $f:(X,\tau)\to (Y,\rho)$ is an embedding with dense image and $S\subseteq X$, then $S$ is nowhere dense if and only if $f[S]$ is nowhere dense in $Y$.

Consequently, so is being first category. Since the irrationals are not first category in $\Bbb R$, they aren't in their subspace topology either.

To prove the central assertion, note that $\operatorname{cl}_XS$ contains a non-empty open set if and only if there is some $U\in\rho$ such that $\operatorname{cl}_{f[X]}f[S]\supseteq U\cap f[X]\supsetneq \emptyset$. If the latter, then $$\operatorname{cl}_Yf[S]= \operatorname{cl}_Y\operatorname{cl}_{f[X]}f[S]\supseteq\operatorname{cl}_Y(U\cap f[X])\stackrel{\text{density of }f[X]}\supseteq U$$ and therefore $f[S]$ is not nowhere dense in $Y$.

On the other hand, if $f[S]$ is not nowhere dense in $Y$, then $\operatorname{cl}_Yf[S]$ contains a non-empty open set $U$, and therefore $$\operatorname{cl}_{f[X]}f[S]=f[X]\cap \operatorname{cl}_Yf[S]\supseteq U\cap f[X]\stackrel{\text{density}}\ne\emptyset$$