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Theorem Let X be. topological space. A non-emptysubset A of X is connected iff there does not exist open subset s U, and V of X such that A $\cap U\ne\emptyset$ A$\cap V\ne \emptyset$ ,A$\cap U\cap V=\emptyset$ and A$\subset U \cup V$

Attempted proof I have proved it going forward,but am having trouble proving the converse.

<= Suppose otherwise ,so there exists open subsets U and V of X such that neither A $\cap U \ne \emptyset$ ,neither A $\cap V\ne \emptyset$ ,neither A$\cap U\cap V=\emptyset$

If A $\cap U \ne \emptyset$ is false and A $\cap V\ne \emptyset$ is false then both intersections are empty,U and V form a disconnected sets and A is connected?

Any help would be appreciated. I notice in MSE and elsewhere this is a definition . My text A first course in topology by Conover,he gives it as a theorem

I can only use contradiction method. Thanks

  • What defintion has been for connectedness of a set? – PCeltide Nov 03 '21 at 19:33
  • A topological space X is connected if it can’t be written as union of two non-empty disjoint open sets –  Nov 03 '21 at 20:43
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    If there exist $U$, $V$ open in $X$ with the listed properties, then $A\cap U$ and $A\cap V$ will form a separation of $A$. – Yinfeng LU Nov 04 '21 at 02:00
  • I redid my proof based off this URL: https://math.stackexchange.com/q/2784131/960654 would it work? But haven’t entered it in my notebook yet as accepi@Yinfeng LU –  Nov 04 '21 at 13:07

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Sorry, my previous comment proves the forward direction, not the converse you are asking for. To prove the converse, we suppose $A$ is not connected, then $A$ has a separation: $$A=H\cup K,\quad\text{where }H,K\text{ open in }A,\quad H\neq\emptyset,\quad K\neq\emptyset,\quad H\cap K=\emptyset.$$ But $H$ open in $A$ means $H$ is the intersection of $A$ with an open subset of $X$ (by the definition of subspace topology), say $H=A\cap U$. Similarly, $K=A\cap V$ for some $V$ open in $X$. Then clearly $U$ and $V$ satisfy $A\cap U=H\neq\emptyset$, $A\cap V=K\neq\emptyset$, $A\cap U\cap V=(A\cap U)\cap(A\cap V)=H\cap K=\emptyset$, and $A=H\cup K\subset U\cup V$. Hence, there do exist $U$ and $V$ open in $X$ with the listed properties.

Taking the contrapositive gives you the converse.

Yinfeng LU
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