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I'm having some trouble proving the following:

If $(X,\tau)$ is a Hausdorff door space, then at most one point $x \in X$ is a limit point of $X$.

My approach was the following:

I assumed that there existed $y \in X$ such that $y$ is also a limit point of $X$, and then prove by contradiction that if $x$ is already a limit point, then the point $y$ cannot exist. However, I'm having some trouble completing this proof. Is this the correct approach? Any tip on how to solve this?

Eduardo Magalhães
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  • what is a door space? – Cronus Aug 13 '20 at 13:41
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    A topological space $(X,\tau)$ is said to be a door space if every subset $S$ of $(X,\tau)$ is either open or closed (or both) @Cronus – Eduardo Magalhães Aug 13 '20 at 13:43
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    Hope you don't mind. I have edited your title so it is more concise and we can use it as a reference when adding the door property to https://topology.pi-base.org/properties. (does not detract anything from your question, as the text itself spells out the theorem again) – PatrickR May 07 '23 at 05:17

2 Answers2

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Suppose $x$ and $y$ are two distinct limit points. Let $x\in U$ and $y\in V$ be separating disjoint open sets for $x$ and $y$. Then $(U-\{x\})\cup \{y\}$ is not open because it contains $y$ but no other points of $V$, and it is not closed because its closure contains $x$. This contradicts the fact that $X$ is a door space.

Matt Samuel
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  • You said that "$(U - {x}) \cup {y}$ is not open because it contains $y$ but no other points of $V$". Why does that imply that "$(U - {x}) \cup {y}$ is not open"? – Eduardo Magalhães Aug 13 '20 at 14:12
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    @EduardoMagalhães If it were open, there would be an open set $A$ with $y\in A$ and $A\subset (U-{x})\cup {y}$. The set ${y}$ is not open because every open set containing $y$ contains another point, and hence $A\cap V$ contains $y$ and some other point of $V$. There is therefore no such $A$, so the set can't be open. – Matt Samuel Aug 13 '20 at 14:14
  • How can you conclude that if $(U - {x}) \cup {y}$ were open, then it would exist a open set $A$ such that $y \in A$ and $A \subset (U - {x}) \cup {y}$? – Eduardo Magalhães Aug 13 '20 at 15:30
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    @EduardoMagalhães Well you could even take $A=(U-{x})\cup{y}$, then $A\cap V$ is nonempty and open and contains $y$ hence it contains an additional point of $V$. – Matt Samuel Aug 13 '20 at 15:32
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    @EduardoMagalhães We showed $A=U-{x}\cup{y}$ is open, and it contains $y$. This means it must contain some other point from $V$, or else $A\cap V={y}$, and this would mean ${y}$ is open, and hence not a limit point, which contradicts our assumptions. – Cronus Aug 17 '20 at 04:51
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To complement Matt Samuel's answer, the following seems an interesting corollary to state.

Every Hausdorff door space is scattered.

where a space $X$ is scattered if every nonempty set $A\subseteq X$ has a point isolated in $A$.

Proof: If $A$ is a singleton, it has an isolated point. And if $|A|\ge 2$, by the main result $A$ contains at least one point isolated in $X$. That point is also isolated in $A$.

PatrickR
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