Projection from ordered square is closed

Question

Consider $I = [0,1]$ and the topology induced by the lexicographic order on $I^2$. Let $\pi$ be the projection between $I^2$ and $I$ that sends the tuple $(x,y) \mapsto x$, where $I$ has the usual topology. Is $\pi$ a closed map? I have demonstrated that $\pi$ it is continous. My only idea is to show that $I^2$ is compact so the closed of $I^2$ must be compacts also and compactness is preserved by continous maps. However, I would prefer to show the property directly since the proofs I have seen of compactness are quite involved. Thank you in advance.

Answer 1

Suppose that $C\subseteq I^2$ is closed, for convenience let $A=\pi[C]$, and fix $x\in\operatorname{cl}_IA$. There is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ converging to $x$, and for each $n\in\Bbb N$ there is a $y_n\in I$ such that $\langle x_n,y_n\rangle\in C$. $I$ is compact, so the sequence $\langle y_n:n\in\Bbb N\rangle$ has a convergent subsequence, and without loss of generality we may assume that $\langle y_n:n\in\Bbb N\rangle$ converges to some $y\in I$. But then $\big\langle\langle x_n,y_n\rangle:n\in\Bbb N\big\rangle$ converges to $\langle x,y\rangle$, and since $C$ is closed, we must have $\langle x,y\rangle\in C$ and hence $x\in A$. Thus, $A$ is closed, and therefore so is the map $\pi$.