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If $E$ is a Hausdorff space, and $A\subset E$ how to prove that $\left(\overline{A}\right)'\subset A'$ ?

We say that $x\in A'$ if and only if $\forall V\in \mathcal{V}_x, (V\setminus\{x\})\cap A\neq\emptyset$

We say that $x\in \overline{A}$ if and only if $\forall V\in \mathcal{V}_x, V\cap A\neq\emptyset$

So let $x\in \left(\overline{A}\right)' $ then $\forall V\in \mathcal{V}_x, (V\setminus\{x\})\cap \overline{A}\neq \emptyset$

then there exist $y$ such that $y\in V\setminus\{x\}$ and $y\in \overline{A}$ so $y\in V$ and $\forall W\in \mathcal{V}_y, W\cap A\neq \emptyset$

We have that $E$ is a Hausdorff space so there exist a nbh $V_1$ to $x$ and $V_2$ for $y$ such that $V_1\cap V_2= \emptyset$

But how we prove that $(V\setminus\{x\})\cap A\neq \emptyset$

how to continue please

Vrouvrou
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3 Answers3

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Note that $\overline A = A \cup A'$. Since the derived set of a finite union is the union of the derived sets, we have $\left(\overline A\right)' = A' \cup A''$. Since $A'$ is closed in a $T_1$-space (see Set of limit points of a subset of a Hausdorff space is closed.), we have $A''\subseteq A'$ and thus $\left(\overline A\right)'\subseteq A'$.

Community
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Stefan Hamcke
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Perhaps it is easier to prove the contrapositive: $(A')^c \subset ((\overline{A})')^c$.

Suppose that $x \in (A')^c$, so $x$ is not a limit point of $A$. Therefore there exists $V \in \mathcal V_x$ such that $W = V \setminus \{x\}$ does not intersect $A$. If we can show that $W$ does not intersect $\overline A$ then we're done.

We already know that $W$ does not intersect $A$, so it suffices to show that $W$ does not contain any limit points of $A$.

Note that $\{x\}$ is closed since the space is Hausdorff (see proof below), so $W = V \setminus \{x\} = V \cap (\{x\})^c$ is an intersection of two open sets, hence open. Suppose $y \in W$. If $y$ were a limit point of $A$, then $W$ would have to contain some $a \in A$, but this cannot happen because $W \cap A = \emptyset$.

Proof that all singletons are closed in a Hausdorff space: let $\{x\}$ be any singleton and $y$ be any other point. Then there is an open neighborhood $U_y$ which contains $y$ but does not contain $x$. Then $$U = \bigcup_{y \in E \setminus \{x\}} U_y$$ is a union of open sets, hence open, and $U$ contains every point of $E$ except $x$, hence $U = \{x\}^c$. Since $\{x\}^c$ is open, it follows that $\{x\}$ is closed.
  • please why $W$ is open ? – Vrouvrou Nov 05 '15 at 19:48
  • @Vrouvrou: In a Hausdorff space, all singletons are open, so ${x}$ is an open set. Therefore, $W = V \setminus {x} = V \cap ({x})^c$ is the intersection of two open sets, hence $W$ is open. –  Nov 05 '15 at 19:57
  • @Vrouvrou: I added a proof that singletons are closed in a Hausdorff space (or more generally, in any $T1$ space, which is a space where if $x$ and $y$ are disjoint, then there is an open neighborhood of $x$ which does not contain $y$, and vice versa; this is of course true in a Hausdorff space since we can find disjoint open neighborhoods of $x$ and $y$). –  Nov 05 '15 at 20:07
  • So the main result holds in any $T_1$ space. – DanielWainfleet Nov 06 '15 at 08:54
  • I don't understand $W$ can have similar points with $\overline{A}$ where this points are not in $A$ – Vrouvrou Nov 06 '15 at 13:21
  • @Bungo please i don't understand your proof – Vrouvrou Nov 06 '15 at 14:09
  • @Vrouvrou:I'm not sure I understand your question. $W$ is defined so it does not intersect $A$, but $\overline A$ is larger than $A$, so we need to verify that $W$ does not intersect $\overline A$. –  Nov 06 '15 at 17:17
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Let $x\in\left(\overline{A}\right)'$ and let $V$ be an open set with $x\in V$.

Then $(V\setminus\{x\})\cap\overline{A}\neq\varnothing$. Let $y\in(V\setminus\{x\})\cap\overline{A}$.

In order to prove that $x\in A'$ it is enough to show that $(V\setminus\{x\})\cap A\neq\varnothing$.

We will do that by deducing a contradiction on base of the assumption: $(V\setminus\{x\})\cap A=\varnothing$.

Note that $V$ is an open set that contains $y\in\overline{A}$, so that $V\cap A\neq\varnothing$.

$(V\setminus\{x\})\cap A=\varnothing$ combined with $V\cap A\neq\varnothing$ implies $V\cap A=\{x\}$.

We have $y\in V\setminus\{x\}$ so that $y\neq x$.

Let $U$ be an open set with $y\in U$ and $x\notin U$. Such set exists because the space is Hausdorff.

(Actually it is enough allready if the space is T1.)

Then $y\in U\cap V$, $U\cap V$ is open and $(U\cap V)\cap A=\varnothing$.

This contradicts that $y\in\overline{A}$ and we are ready.

drhab
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  • please why you take the same $V$ you must say that there exist $V_1\in \mathcal{V}_x$ such that $V_1\setminus{x})\cap A=\emptyset$ – Vrouvrou Nov 06 '15 at 14:03
  • I am afraid that I do not understand you (especially the words "the same"). I start with on open set $V$ that contains $x$. In your notation probably $V\in\mathcal V_x$. Then it is proved that the assumption $(V\setminus{x})\cap A=\varnothing$ leads to a contradiction. Conclusion: $(V\setminus{x})\cap A\neq\varnothing$. This for every $V\in\mathcal V_x$ so that $x\in A'$. There is no need to refer to some (other?) $V_1\in\mathcal V_x$. – drhab Nov 06 '15 at 14:13
  • It might be that "my" $V$ (neighborhood of $x$) corresponds with "your" $V_1$, and that "my" $U$ (neighborhood of $y$) corresponds with "your" $V_2$. – drhab Nov 06 '15 at 14:27
  • I can say this:assume that $x\in \left(\overline{A}\right)'$ and $x\notin A'$ i.e $\forall V\in \mathcal{V}_x, V\setminus{x} \cap \overline{A}\neq \emptyset....(1)$ and $\exists V_1\in \mathcal{V}_x, V_1\setminus{x}\cap A=\emptyset$ as $(1)$ i true for all $V$ then it is is true for $V_1$ so $V_1\setminus{x}\cap \overline{A}\neq\emptyset$ so $\exists y\in V_1\setminus{x}\cap \overline{A}$ and find a contradition with $y\in \overline{A}$ – Vrouvrou Nov 06 '15 at 15:02
  • Do you understand my proof? If not, then what is unclear to you? – drhab Nov 06 '15 at 15:08
  • i don't understand why you say that $V\setminus{x}\cap A=\emptyset$ we mut fix $V$ because $x\notin A'$ means that there exist V not for all V – Vrouvrou Nov 06 '15 at 15:13
  • Actually I am not saying that $V\setminus{x}\cap A=\varnothing$, but I am assuming that $V\setminus{x}\cap A=\varnothing$. Based on that assumption I reduce a contradiction, wich allows me to conclude that the assumption is false, so that $V\setminus{x}\cap A\neq\varnothing$ must be true. Exactly what I am after. This is proved for an arbitrary (open) neighborhood of $x$, so it works for any neighborhood of $x$. – drhab Nov 06 '15 at 15:17
  • Let us continue this discussion in chat. – Vrouvrou Nov 06 '15 at 15:25
  • I am waiting for you in the chatroom. – drhab Nov 06 '15 at 15:28
  • I have been waiting in vain. – drhab Nov 06 '15 at 15:37