Can anyone give me a proof of the statement that if $X$ and its closure $\overline X$ are connected and if $X\subset Y \subset \overline X$, show that Y is also connected?
Thank you.
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5The hypothesis that $\overline X$ be connected is not necessary. – David Mitra Jan 22 '13 at 17:43
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@DavidMitra yeah, The closure of a connected set is connected. – M.Sina Jan 22 '13 at 17:45
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Related – rschwieb Jan 22 '13 at 17:59
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In case you are using the "you can't write it as a union of disjoint open sets" definition, like I grew up with:
Let $O_1$ and $O_2$ be open sets of the universal set whose union contains $Y$ and such that $(O_1\cap Y)\cap(O_2\cap Y)=\emptyset$.
Since their union also contains $X$, which is connected, one of them, say $O_1$ is such that $X\cap O_1=\emptyset$ (and $X\subseteq O_2$).
But then by definition of the closure, $\overline{X}\cap O_1=\emptyset$, and hence $Y\cap O_1=\emptyset$. So, $Y\subset O_2$.
rschwieb
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@DavidMitra Ah thanks: yes, I should make that correction. Glad you helped! – rschwieb Jan 22 '13 at 19:58
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Why $X\cap O_{1}=\emptyset \implies \overline {X} \cap O_{1} = \emptyset$ ? – Astro Nauft Feb 04 '15 at 14:58
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@AstroNauft That is the definition of closure: any open set touching $\overline{X}$ must touch $X$. – rschwieb Feb 04 '15 at 16:57
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@rschwieb Sorry so what is the conclusion here exactly? So you showed that $Y \cap O_1$ empty, and $Y \cap O_2$ is $Y$, then $\varnothing \cap Y = \varnothing$ where is the contradiction? – Olórin Feb 08 '16 at 01:15
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@EmotionallyVulnerableLlama the conclusion is that $Y$ cannot be disconnected with open sets (there is no contradiction because it is not a proof by contradiction) So, Y is connected – rschwieb Feb 08 '16 at 02:18
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@rschwieb Hi sorry for being so dumb but why wouldn't $Y$ be disconnected within $O_2$? – Olórin Feb 08 '16 at 02:35
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@EmotionallyVulnerableLlama I think if you stop and think carefully about how this proof works, you'll realize you are distracting yourself. I am showing that all disconnections of Y are trivial. That is, if Y is contained in the union of two open sets whose intersection in Y is empty, then one of them already contains Y. A nontrivial disconnect would be a pair of such sets which both contain points of Y. – rschwieb Feb 08 '16 at 04:50
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If $Y$ is not connected, $Y$ is union of nonempty seperated set $A,B$. Since $X$ is connected, Then $X\subset A$ or $X\subset B$.
If $X\subset A$, then $\overline{X}\subset \overline{A}$. So $Y\subset \overline{A}$ But since $A$, $B$ is seperated, $\overline{A}\cap B$ is empty and $B\subset \overline{A}$, so $B=\varnothing$, which derive a contradiction.
Hanul Jeon
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@Seirios correct some expression of this proof.
Because $Y\subset \overline{A}$ and $Y=A\cup B$.
– Hanul Jeon Jan 22 '13 at 17:17 -
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1@Jack You can find the equivalent conditions of disconnectedness - $A$ and $B$ are separated if and only if $\overline{A}\cap B$ and $A\cap\overline{B}$ are empty. – Hanul Jeon Sep 29 '15 at 14:24
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Let $g : Y \to \{0,1\}$ be a continuous function. Then $g_{|X} : X \to \{0,1\}$ is also continuous, and because $X$ is connected, $g$ is constant on $X$; because $X$ is dense in $Y$, you deduce that $g$ is constant on $Y$.
Therefore, every continuous function from $Y$ to $\{0,1\}$ is constant: $Y$ is connected.
Seirios
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Shouldn't $Y$ be Hausdorff to deduce $g$ is constant on $Y$? – Camilo Arosemena Serrato Jan 22 '13 at 16:54
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2@CamiloArosemena: Let $x \in \overline{X} \cap Y$. Because $g$ is continuous, there is a neighborhood $V$ of $x$ such that $|g(x)-g(y)|<1/2$ if $y \in V$; but $V$ contains an element $y_0 \in X$, hence $g(x)=g(y_0)$. I think that Hausdorffness is not required here. – Seirios Jan 22 '13 at 17:20
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@CamiloArosemena-Serrato Hausdorffness of the codomain ${0,1}$ is needed yes, but that's OK here as we have a discrete space. The domain's Hausdorffness is irrelevant. – Henno Brandsma Jun 12 '19 at 21:44