equivalent formulation of the tube lemma

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

(TL) If UX×Y is open (in product topology) and if xX is such that x×YU, then there exists an open neighbourhood VX of x such that V×YU.

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

(CP) The projection π:X×YX given by π(x,y)=x is a closed map.

The following theorem relates these two statements:

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

Proof. ,,” Let FX×Y be a closed setPlanetmathPlanetmath and let U=(X×Y)\F be its open complementPlanetmathPlanetmath. We will show, that π(F) is closed, by showing that V=X\π(F) is open. So assume, that xV. Obviously


Therefore x×YU and by (TL) there exists open neighbourhood VX of x such that V×YU. It easily follows, that VV and it is open, so (since x was chosen arbitrary) V is open.

,,” Let UX×Y be an open subset such that x×YU for some xX. Let F=(X×Y)\U. Then F is closed and by (CP) we have that π(F)X is closed. Also xπ(F) and thus V=X\π(F) is an open neighbourhood of x. It can be easily checked, that V×YU, which completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

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 HausdorffPlanetmathPlanetmath with Y compactPlanetmathPlanetmath, then both are true. On the other hand, for example for X=Y=, where denotes reals with standard topology, they are both false. For example consider


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

Title equivalent formulation of the tube lemma
Canonical name EquivalentFormulationOfTheTubeLemma
Date of creation 2013-03-22 19:15:18
Last modified on 2013-03-22 19:15:18
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Theorem
Classification msc 54D30