So I thought that I understood this, but now I am getting scrambled. A set is closed if it contains its limit points. I thought that that meant that any point in a closed set would contain the points in its neighborhood. But I was looking at the proof for $E'$ being closed, and it was stating that a point in $E'$ would have a neighborhood with a point in $E$, which I thought would mean that $E'$ is not closed or that $E'= E$, which won't always be the case. What am I missing here? Thanks!
Topology: $E'$ is Closed
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real-analysis
general-topology
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0I have already posted an answer, but rereading the question, I think I missed the source of your confusion. Typically, $E'$ denotes the set $E$ along with all of its limit points, so $E'$ is *always* closed, by definition. Is that what $E'$ means to you? – 2017-02-26
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0@TheCount $E'$ never denotes that. It's always the set of limit points. And $E \cup E'$ is called the closure $\overline{E}$ of $E$. Its closedness requires a small argument. – 2017-02-27
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0Ah, yes @HennoBrandsma. T'was late. Silly error on my part. – 2017-02-27
3 Answers
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Not all sets have limit points. Consider $\mathbb{N}$, for example. It is closed because it contains all its limit points (since it has none). Does that help?
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One definition of the limit point of a set is: $x\in X$ is a limit point of the set $E\subseteq X$ if there is an open neighborhood $U$ of $x$ s.t. $(U\setminus \{x\})\cap E\neq \emptyset$ i.e. $x\in X$ is a limit point of the set $E$ if it would have a neighborhood with a point in $E$.
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Some ramblings:
A limit point $x$ of $E$ is one that for every open set $O$ that contains $x$, we know that $O \cap (E\setminus \{x\}) \neq \emptyset$, so every neighbourhood of $x$ contains a point of $E$ different from $x$ itself. The last clause is because if $x \in E$, then every open ball around $x$ intersects $E$ in $x$ itself, so $x$ but this might have been the only point form some open neighbourhood. This $x$ is then an isolated point of $E$.
In a metric space (or first countable space), a limit point of $E$ is one that is the limit of a sequence of points from $E$ ,all diffferent from $x$. This "explains" the name a bit. In any space $X$, points of $X$ can either be isolated points of $E$, interior points of $X \setminus E $ (points in the exterior of $E$) or limit points of $E$. A set like $\mathbb{Z}$ does not have any limit points, all pointsof it are isolated. The set $E = \{\frac{1}{n}: n=1,2, \ldots\}$ has only $0$ as a limit point, all points of $E$ are isolated. So here $E'$ is not a subset of $E$, there is a limit point of $E$ outside it.
So in a proof that $E'$ is closed (which needs an asssumption on $X$ like $X$ is $T_1$: all finite subsets are closed, as easily holds in all metric spaces) we can start with: suppose $x$ is a limit point of $E'$ (so $x \in E''$) and show it is in $E'$. Then $E'$ contains its limit points and so is closed.
So suppose $x \in E''$. Then let $O$ be any open neighbourhood of $x$. We must show it intersects $E\setminus \{x\}$ to see it is in $E'$. Then $O$ intersects $E' \setminus \{x\}$, say $y \in E', y \in O, y\neq x$. But then as $\{x\}$ is closed, being finite $O \cap X\setminus \{x\}$ is also an open neighbourhood of $y$ and as $y \in E'$, $O \cap (X\setminus \{x\}) \cap E \setminus \{y\} \neq \emptyset$, so in particular $O \cap E \setminus \{x\} \neq \emptyset$. So $x \in E'$ and so indeed $E'' \subset E'$. (Note that this doesn't say that $E''$ is non-empty, as the proof of an inclusion $A \subset B$ only states that if we have a point of $A$, it must also b in $B$. If there are no points in $A$, this is automatically true, so $\emptyset \subset B$ regardless of $B$, similarly $E' \subset E$ is true for $E$ with no limit points at all, but the proof I just gave works regardless of emptyness of $E'$ or $E''$).