Showing that a topological space is connected iff for every two points, there is a connected subspace that contains them.

general-topologysolution-verification

Here is my attempted proof of the proposition provided below. My question is: is this proof attempt valid, and furthermore, can it be improved?

Proposition. A topological space $X$ is connected iff for any two points $x, y \in X$, there exists a connected subspace $U \subseteq X$ such that $x \in U$ and $y \in U$.

I will take the following fact as given.

Lemma 1. A topological space $X$ is connected iff it has no non-trivial clopen subsets i.e. a clopen subset of $X$ is either empty $\emptyset$ or $X$.

Proof attempt. The forwards direction is trivial: given any connected space $X$, $X$ is a connected subspace of itself that contains any two points $x, y \in X$. For the converse, let $X$ be a space such that for any $x, y \in X$, there exists a connected subspace $U \subseteq X$ such that $x \in U$ and $y \in U$. We need to show that $X$ is connected. By Lemma 1, it suffices to show that no subset of $X$ is non-trivial and clopen. Suppose towards a contradiction that there exists some $A \subseteq X$ that is non-trivial and clopen. As $A$ is non-trivial, both $A$ and $A^c$ must be inhabited i.e. there must exist some
\begin{equation*}
x \in A \qquad \text{and} \qquad y \in A^c.
\end{equation*}
Notice that there must exist a connected subspace $U \subseteq X$ such that $x \in U$ and $y \in U$. As $U$ is a connected subspace, it must have no non-trivial clopen subsets (by Lemma 1). Consider, however, the set $A \cap U$ which must be clopen (in the subspace topology): it must be open as it is the intersection of an open set of $X$ with $U$ and it must be closed as its complement (with respect to $U$), $A^c \cap U$, is open as $A^c$ is open. We know that $A \cap U$ is also a non-trivial subset of $U$ since $x \in A \cap U$ (as $x \in A$ and $x \in U$) and $y \notin A \cap U$ (as $y \in A^c$).

Best Answer

Yes, the proof is fine. An alternative proof, assuming you know this lemma

Lemma If $\{C_i\mid i \in I\}$ is a set of connected subspaces of $X$ and $\bigcap_i C_i \neq \emptyset$ then $\bigcup_i C_i$ is connected.

Then fix $x \in X$ and for each $ y \neq x$ let $C(x,y)$ be a connected subspace of $X$ that contains $x$ and $y$. We apply the lemma to $\{C(x,y): y \neq x\}$ and note that $x$ is in their intersection, and $X= \bigcup_{y \neq x} C(x,y)$ is connected.

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