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Let $X$ be a set and $\{U_{\alpha}\}$ is a collection of subsets of X. Suppose the topology on each $U_{\alpha}$ is defined and each of them is path connected. Then define a topology on $X$ by declaring that a subset $U\subset X$ is open in $X$ if $U\cap U_{\alpha}$ is open in $U_\alpha$ for each $\alpha$.

Now, form a graph with one vertex(called $v_\alpha$) for each $U_\alpha$ and with each vertex $v_\alpha$ connected by an edge to $v_\beta$ if and only if $U_\alpha \cap U_\beta \ne \varnothing $.(what's the terminology for this graph?)

I believe that if the graph is connected and $X=\cup_\alpha U_\alpha$ then $X$ is path connected. But is it also true that when $X$ is path connected, then the graph is also connected?(I am not very sure if this direction also needs the condition that $\cup_\alpha U_\alpha=X$)

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    I don't know what the graph is called. But indeed, the space is path-connected if and only if the graph is connected. – 2017-01-21
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    The graph is often called the nerve of the cover – 2017-01-21
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    @DanielFischer Then how to prove the direction I asked? – 2017-01-21
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    The nerve also has simplicies of higher dimension. This would be the 1-skeleton of the nerve of the covering. – 2017-01-21
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    @DanRust He's giving $X$ the coherent topology, so the $U_\alpha$s are open. – 2017-01-21
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    Are you assuming that the $U_\alpha$s cover $X$? – 2017-01-21
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    @TiWen Take $\alpha,\beta$, and pick $x\in U_{\alpha}$ and $y\in U_{\beta}$. Take a path $\gamma\colon [0,1] \to X$ connecting $x$ and $y$. $\{U_{\iota}\}$ is an open cover of $\gamma([0,1])$, pick a finite subcover (containing $U_{\alpha}$ and $U_{\beta}$). If $\gamma([0,1]) \subset U_{\alpha}$, we're done, then $y\in U_{\alpha}\cap U_{\beta}$, so there is even an edge between $v_{\alpha}$ and $v_{\beta}$. Otherwise, let $t_1 = \min \{ t \in [0,1] : \gamma(t) \notin U_{\alpha}\}$. Then $\gamma(t_1) \in U_{\iota_1}$, – 2017-01-21
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    and by continuity of $\gamma$ we have $\gamma(t) \in U_{\iota_1} \cap U_{\alpha}$ for $t \in (t_1-\varepsilon, t_1)$. So we have an edge between $v_{\alpha}$ and $v_{\iota_1}$. If $\gamma([t_1,1]) \subset U_{\iota_1}$, done again, otherwise let $t_2 = \min \{ t \in [t_1,1] : \gamma(t)\notin U_{\iota_1}\}$. And so on, after finitely many steps you've reached $y\in U_{\beta}$. – 2017-01-21
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    But Alex Provost makes a good point, that is only valid if the $U_{\alpha}$ cover $X$. The graph only tells us anything about $\bigcup U_{\alpha}$. – 2017-01-21
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    @AlexProvost Thank you! Yes, I think both directions need the condition that $U_\alpha$'s cover $X$. – 2017-01-23

2 Answers 2

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Yes, assuming that none of the $U_{\alpha}$ is empty. To show that there is a connection between $v_{\alpha}$ and $v_{\beta}$ in the graph, consider a path $p:[0,1]\to X$ that connects some arbitrary point $x_{\alpha}\in U_{\alpha}$ with some point $x_{\beta}\in U_{\beta}$. $[0,1]$ is covered by the preimages $p^{-1}(U_{\gamma})$, where $\gamma$ rotates through every element of the index set. Every one of these preimages can be written as a union of intervals open in $[0,1]$. Because of its compactness, $[0,1]$ is covered by finitely many of these intervals. If among these there is an interval that is a subset of one of the others, remove it. Repeat this step until none such interval remains. Thus we have indices $\gamma_0,\gamma_1,..,\gamma_k$ with wlog $\gamma_0 = \alpha$, $\gamma_k=\beta$, intervals $I_{\gamma_j}$ open in $[0,1]$, so that $I_{\alpha} = [0,t_1)$, $I_{\beta}=(t_{k},1]$, $I_{\gamma_j}= (t_j,t_{j+1})$ for $10$, showing that there is an edge between $v_{\gamma_j}$ and $v_{\gamma_{j+1}}$ in the graph. Thus $v_{\alpha}=v_{\gamma_0}$ is connected to $v_{\beta}=v_{\gamma_k}$ in the graph.

If one of the $U_{\alpha}$ is empty, then $v_{\alpha}$ is connected to no other node.

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    I would say you don't need to assume the $U_\alpha$ are empty because the empty set is not path-connected. – 2017-01-21
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    Oh, is that standard definition? If so, my bad. I thought it's just the "for all" property. – 2017-01-21
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    I think the empty set is normally taken to be both connected and path connected, although many authors will differ on convention. – 2017-01-21
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Suppose the graph is nonempty and disconnected; say you can partition the $v_\alpha$ into two nonempty sets $A$ and $B$ with no edges between them. Let $U=\bigcup_{\alpha\in A} U_\alpha\cup X\setminus\left(\bigcup_{\alpha} U_\alpha\right)$ and $V=\bigcup_{\beta\in B}U_\beta$. Note that for any $\alpha$, either $U_\alpha$ is contained in $U$ (if $\alpha\in A$) or disjoint from $U$ (if $\alpha\in B$, since there are no edges between $A$ and $B$), so $U$ is open. Similarly, $V$ is also open. Since there are no edges between $A$ and $B$, $U\cap V=\emptyset$, and since every $\alpha$ is in either $A$ or $B$, $U\cup V=X$. Thus $U$ and $V$ witness that $X$ is disconnected.

(If the graph is empty, then $X$ has the discrete topology, so $X$ will still not be connected unless it has exactly one point.)

(Note that the converse which you claimed to already know is only true if the $U_\alpha$ cover $X$. For if they don't cover $X$ (and there exists at least one $U_\alpha$), then the union of all the $U_\alpha$ is a nontrivial clopen subset of $X$, regardless of whether the graph is connected.)

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    How is $X$ connected? The set $U_1$ and its complement are both open in $X$. Or more simply, if you let $x$ be any point not contained in $U_1$ or $U_2$, then $\{x\}$ and its complement are both open. – 2017-01-23
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    You are right, that $X=\mathbb R$ example is not endowed with usual topology. Now I think Your solution is better. – 2017-01-23