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This is a question I had, while trying to solve a homework problem. My original approach was dependent upon the following statement being true.

If $Y$ and $Y \cup X$ are connected, then there is some connected component of $X$, call it $C$, where $Y \cup C$ is connected.

I eventually solved the homework problem using a very different method, but the question has been bugging me. I can't seem to come up with a proof or counterexample.

I haven't been able to make more than trivial progress (for example, if $Y$ is not closed or there is some component $C$ is not closed. Then $\overline{C} \cap \overline{Y} \neq \emptyset$ and you can show that one must contain a limit point of the other, therefore $Y \cup C$ is connected.)

Does anyone have any insights?

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  • $\begingroup$ What happens if $X$ is not a connected set? $\endgroup$
    – abiessu
    Nov 14, 2013 at 0:36
  • $\begingroup$ $Y \cup X$ are connected? That doesn't make sense. (Perhaps setting $A = Y \cup X$ and saying $A$ is connected is better.) $\endgroup$
    – Don Larynx
    Nov 14, 2013 at 0:37
  • $\begingroup$ @DonLarynx ($Y \cup X$) and $Y$ are both connected. Does that make more sense? $\endgroup$
    – Bill Trok
    Nov 14, 2013 at 0:40

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The Knaster-Kuratowski fan is a counterexample: let $Y$ be the apex and $X$ the rest of the space. Now, $Y$ is a point and thus connected, and so is $Y\cup X$. But components of $X$ are individual points. So, $Y\cup C$ is disconnected for every component $C$ of $X$, as it is a two point Hausdorff space.

(I imagine there should also be simpler counterexamples. If I think of one, I'll let you know.)

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  • $\begingroup$ Is my answer similar to yours? (Not implying anything else, just wondering) $\endgroup$
    – Don Larynx
    Nov 14, 2013 at 0:56
  • $\begingroup$ @DonLarynx: nope, my answer claims that there is a counterexample, whereas your answer claims that there is a proof. $\endgroup$
    – Dejan Govc
    Nov 14, 2013 at 0:58
  • $\begingroup$ But do we both agree on the same thing? (and I'd actually like feedback on my proof if you don't mind, I'm a student) $\endgroup$
    – Don Larynx
    Nov 14, 2013 at 0:59
  • $\begingroup$ @DonLarynx: actually, we disagree. I'll comment on your proof in a moment. $\endgroup$
    – Dejan Govc
    Nov 14, 2013 at 1:02
  • $\begingroup$ @DejanGovc Thank you this is exactly what I was looking for. $\endgroup$
    – Bill Trok
    Nov 14, 2013 at 1:17
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There certainly is some connected component of $X$ where $Y \cup C$ is connected. If there wasn't, then you would get a disjoint union of $X$ and $Y$, a contradiction..

Proof. Choose an arbitrary $x$ in $C$ with the restriction it is the closest point to $Y$ (for some point $y \in Y$). For it to be connected there would have to exist an $\epsilon > 0$ such that $d(x, y) < \epsilon$. If there were no such component, that is, if $d(x, y) > \epsilon$ for all $x$ in $C$ (and since $C$ is an arbitrary subset of $X$, we can just claim $x \in X$ WLOG) then $X$ and $Y$ are clearly disconnected.

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    $\begingroup$ would you mind elaborating on that a bit? That is not true in general for arbitary components, what you would get is a disjoint union of $Y \cup C$, which is entirely possible. $\endgroup$
    – Bill Trok
    Nov 14, 2013 at 0:41
  • $\begingroup$ I pose my question again... What if $X$ is not connected? It doesn't mean that $Y\cup C$ cannot be connected, but a "connected component of $X$" couldn't exist... $\endgroup$
    – abiessu
    Nov 14, 2013 at 0:44
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    $\begingroup$ @abiessu Connected components of $X$ can very well exists, in fact every point of $X$ must be contained in a connected component. Look at the topology inherited by $[0,1]\cup [2,3]$ is not a connected subset, but it has connected component $[0,1]$. $\endgroup$
    – Bill Trok
    Nov 14, 2013 at 0:50
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    $\begingroup$ @abiessu Your statement "A connected component of $X$ couldn't exist", doesn't make any sense all topological sets have connected components, every single one. $\endgroup$
    – Bill Trok
    Nov 14, 2013 at 1:20
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    $\begingroup$ @DonLarynx First of all, you are assuming that we are dealing with metric spaces here. Additionally I would guess that you are talking about the real number line. $\endgroup$
    – Bill Trok
    Nov 14, 2013 at 1:23

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