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I am trying to figure out what the closure of an open interval is in the particular point topology on $\mathbb{R}$ defined by $\mathcal{T}=\{\emptyset\} \cup \{ U\subseteq \mathbb{R} : p \in U \}$.

To make my question easier to write, let's say $p=1$ so every open set in this topology must contain 1 (or be the empty set). Let $A=(-2,-1)$. This interval is not open in this topology since $1\not \in A$. Up to here, I am not confused. My misunderstanding begins when I consider the closure of this set.

Before I write any more regarding my question, below are some definitions I'm working with.

  1. The closure of a set $A$ is the set unioned with its limit points, i.e. $\overline{A}=A\cup A'$.
  2. A point $x\in A$ is a limit point if every open neighborhood, say $U$ of $x$ intersects $A$ at some point other than itself; i.e. $U-\{x\}\cap A\neq \emptyset$.
  3. The closure of a closed set $A$ is simply $A$; i.e. if $A$ is closed in $X$, then $\overline{A}=A$.

Ok now that I've listed what I know, here is what I've tried with the above example. Since $A=(-2,-1)$ is closed in the particular point topology, $\overline{A}=A$. But when I consider the limit points of $A$, I get a different answer. I think the limit points of $A$ are actually $(-\infty, -1 )$ since the open set of any point less than or equal to -2 must include 1. So if I want to check if -3 is a limit point, the open neighborhood of -3 is $(a,b)$ where $a<-3$ and $b>1$. This open neighborhood indeed intersects $A$ at some point other than -3 so -3 must be a limit point of $A$. (And I can make a similar argument for every point between $-\infty$ and -2.)

Clearly, both of those answers cannot be correct so where am I going astray? Am I misunderstanding this topology completely? Are my definitions incorrect?

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Observe that in the particular point topology around $1 \in \mathbb{R}$ we have that $\{x, 1\}$ is open for any $x \in \mathbb{R}$, and this is the "smallest" open set containing $x$. Thus if $x$ is a limit point of $(-2,-1)$ we must have that $(-2,-1) \cap \{x,1\} \neq \emptyset$ which only occurs when $x \in (-2,-1)$. Therefore using the given definition of limit point, we see that $(-2,-1)$ has no limit points and writing $A= (-2,-1)$ we see that $A' = \emptyset$. Therefore $\overline{A} = A \cup A' = A$, hence $(-2,-1)$ is closed.

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  • $\begingroup$ I thought in order for $p$ to be a limit point of $(-2,-1)$ every open neighborhood of $p$ must intersect $(-2,-1)$ at some point other than itself. I don't understand what you mean then by "All that is necessary is that every neighborhood of p intersects (−2,−1) somewhere, even if it is just at p itself". Doesn't that go against the definition of a limit point? That aside, I do think I understand, the closure of (-2,-1) in this topology is (-2,-1). $\endgroup$ – Nidia Mar 20 '15 at 5:05
  • $\begingroup$ @NidiaGonzalinajec I edited my answer. Let me know if it makes sense. $\endgroup$ – Mnifldz Mar 20 '15 at 6:24
  • $\begingroup$ that makes much more sense! Thank you! $\endgroup$ – Nidia Mar 20 '15 at 17:02
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This answer is incorrect. I am only leaving it up b/c the comment on this is very helpful.

I have just caught one mistake in my work. A set in a topological space is closed if and only if its compliment is open. The compliment of $A=(-2,-1) $ is $(-\infty,-2)\cup(-1,\infty)$. The interval $(-1,\infty)$ is an open set in this topology (since it contains 1) but $1\not \in (-\infty,-2)$ so this interval is not open. It follows that $A^c$ is not open so $A$ is not closed. Clearly then, $\overline{A}\neq A$.

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  • $\begingroup$ This is wrong. The interval $(-\infty, -2)$ is certainly not open but the set $(-\infty, -2) \cup (-1,\infty)$ is open. $A$ as you have defined it is closed because its complement is open. $\endgroup$ – Mnifldz Mar 20 '15 at 5:01
  • $\begingroup$ I do see that now actually. I didn't want to delete it after I posted it though. Thank you for pointing this out. $\endgroup$ – Nidia Mar 20 '15 at 5:03

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