A topological space is path connected if and only if the associated graph is connected

Question

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 $\cup_\alpha U_\alpha=X$)

Answer 1

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 $1<j<k$ and $t_{j+1}\in I_{\gamma_{j+1}}$ for all $0\leq j <k$, because consecutive intervals (ordered by their left endpoint) must overlap. Now for every $1\leq j<k$, $t_{j+1}-\varepsilon_j\in I_{\gamma_{j}}\cap I_{\gamma_{j+1}}\subseteq U_{\gamma_j}\cap U_{\gamma_{j+1}}$ for some $\varepsilon_j>0$, 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.

Answer 2

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.)