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Though this sounds like an obvious fact, I am trying to prove directly from the definition of a closed set (in Rudin) that $X$, the entire metric space, is closed. Rudin's definition of closed is that $A \subset X$ is closed if it contains all of its limit points, where $x \in A$ is a limit point if for every $\epsilon > 0$, $B_{\epsilon} (x) \cap (A - \{x\}) \neq \emptyset$.

My first doubt is that $X$ may only have one element. If that is the case, $X$ must have no limit points -- in fact, if it's a finite set, it has no limit points, but I don't think that lemma qualifies as "directly from the definition." If I specialize to the case where $X$ is infinite, I might say: suppose $x$ were a limit point of $X$. Given $\epsilon > 0$, there exists $q \in B_{\epsilon} (x) \cap X$ with $q \neq x$. That guarantees $q \in X$, but I don't know how I might deduce $x \in X$. The case that stands out to me is the subspace $\mathbb{Q}$ of $\mathbb{R}$. Irrational numbers are limit points, surely, but they're not elements of $\mathbb{Q}$, so they're not even considered when I'm regarding closedness as a subset of $X$.

Is there a way to prove this fact formally? Is it just "tautological" because $X$ contains everything?

Mathematical Endeavors
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  • If $X$ has no limit points, then it is closed via vacuous logic. (One may think of "it contains all zero limit points.") This is discussed here. – PrincessEev Feb 13 '23 at 20:41
  • Does this answer your question? Is a set without limit points necessarily closed? – PrincessEev Feb 13 '23 at 20:42
  • @PrincessEev It makes sense to me, on reflection, that if $X$ has no limit points, it is vacuously closed. The case where $X$ has a limit point is confusing to me. I'm inclined to start the proof with "let $x$ be a limit point of $X$," implying that $x$ could be "outside" of $X$ (for example, $\sqrt{2}$ is a limit point of the rationals), but I think I don't need to consider points outside of $X$. – Mathematical Endeavors Feb 13 '23 at 20:58
  • Then again, the definition of a limit point may be taken to be "$x \in X$ is a limit point of $A \subset X$....." in which case this condition is in fact vacuous. – Mathematical Endeavors Feb 13 '23 at 21:01
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    $X$ tautologically contains all of its limit points since, by definition, limit points must be points in $X$. When we consider a metric space we consider it "on its own, floating in empty space"; it is not connected to or embedded in any other space. – Qiaochu Yuan Feb 13 '23 at 21:06
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    Just to add a little to Qiauchu Yuan's point, It is not quite true that $\sqrt 2$ is a limit point of the rationals. What is true is that $\sqrt 2$ is a limit point of the rationals as a subset of the real numbers. It is just when we talk about "limit points of the rationals*, the context of being with respect to the real numbers is understood. But when talking about an arbitrary metric space $X$, unless some larger space is explicitly mentioned, all topological operations are with respect to $X$ itself. – Paul Sinclair Feb 13 '23 at 21:24

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