Does path connectedness of $Cl(A)$ implies $A$ is connected?

Question

I recently came across a problem in which it was necessary to prove that $\mathbb Q^2 \cup \mathbb I^2$ is connected but and I went through the path connectivity of $Cl(\mathbb Q^2 \cup \mathbb I^2)$. I heard about this fact from friends, but I still couldn't prove it. Now there are no ideas

So does path connectedness of $Cl(A)$ implies $A$ is connected?

Answer 1

This is of course not true. The closure of $[0,2] \setminus \{1\}$ in the real line is path connected, but the set itself is not connected.

Answer 2

You can say that if $A$ is connected, then any $B$ such that $A\subseteq B\subseteq Cl(A)$ is connected. Also, if $A$ is path connected, then $A$ is connected, hence $Cl(A)$ is connected. But you cannot claim that path connectedness of $Cl(A)$ implies connectedness of $A$, as others have stated.