equivalent formulation of the tube lemma

Let us recall the thesis of the tube lemma. Assume, that $X$ and $Y$ are topological spaces.

(TL) If $U\subseteq X\times Y$ is open (in product topology) and if $x\in X$ is such that $x\times Y\subseteq U$, then there exists an open neighbourhood $V\subseteq X$ of $x$ such that $V\times Y\subseteq U$.

We wish to give a relation between (TL) and the the following thesis, concering closed projections:

(CP) The projection $\pi :X\times Y\to X$ given by $\pi (x,y)=x$ is a closed map.

The following theorem relates these two statements:

Theorem. (TL) is equivalent to (CP).

Proof. ,,$\Rightarrow$” Let $F\subseteq X\times Y$ be a closed set and let $U=(X\times Y)\backslash \backslash F$ be its open complement. We will show, that $\pi (F)$ is closed, by showing that $V=X\backslash \backslash \pi (F)$ is open. So assume, that $x\in V$. Obviously

$$({\pi }^{-1}(x)=x\times Y)\cap F=\mathrm{\varnothing }.$$

Therefore $x\times Y\subseteq U$ and by (TL) there exists open neighbourhood $V^{\prime }\subseteq X$ of $x$ such that $V^{\prime }\times Y\subseteq U$. It easily follows, that $V^{\prime }\subseteq V$ and it is open, so (since $x$ was chosen arbitrary) $V$ is open.

,,$\Leftarrow$” Let $U\subseteq X\times Y$ be an open subset such that $x\times Y\subseteq U$ for some $x\in X$. Let $F=(X\times Y)\backslash \backslash U$. Then $F$ is closed and by (CP) we have that $\pi (F)\subseteq X$ is closed. Also $x\notin \pi (F)$ and thus $V=X\backslash \backslash \pi (F)$ is an open neighbourhood of $x$. It can be easily checked, that $V\times Y\subseteq U$, which completes the proof. $\mathrm{\square }$

Remark. The theorem doesn’t state that any of statements is true. It is well known (see tha parent object), that if both $X$ and $Y$ are Hausdorff with $Y$ compact, then both are true. On the other hand, for example for $X=Y=ℝ$, where $ℝ$ denotes reals with standard topology, they are both false. For example consider

$$F=\{(x,y)\in {ℝ}^2|xy=1\}.$$

Of course $F$ is closed, but $\pi (F)=ℝ\backslash \backslash \{0\}$ is not closed, so the (CP) is false.