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⊆X×Y is open (in product topology) and if x∈X is such that x×Y⊆U, then there exists an open neighbourhood V⊆X of x such that V×Y⊆U.
We wish to give a relation between (TL) and the the following thesis, concering closed projections:
(CP) The projection π:X×Y→X given by π(x,y)=x is a closed map.
The following theorem relates these two statements:
Theorem. (TL) is equivalent to (CP).
Proof. ,,⇒” Let F⊆X×Y be a closed set and let U=(X×Y)\F be its open complement
. We will show, that π(F) is closed, by showing that V=X\π(F) is open. So assume, that x∈V. Obviously
(π-1(x)=x×Y)∩F=∅. |
Therefore x×Y⊆U and by (TL) there exists open neighbourhood V′⊆X of x such that V′×Y⊆U. It easily follows, that V′⊆V and it is open, so (since x was chosen arbitrary) V is open.
,,⇐” Let U⊆X×Y be an open subset such that x×Y⊆U for some x∈X. Let F=(X×Y)\U. Then F is closed and by (CP) we have that π(F)⊆X is closed. Also x∉π(F) and thus V=X\π(F) is an open neighbourhood of x. It can be easily checked, that V×Y⊆U, which completes 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 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)∈ℝ2|xy=1}. |
Of course F is closed, but π(F)=ℝ\{0} is not closed, so the (CP) is false.
Title | equivalent formulation of the tube lemma |
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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 |