equivalent formulation of the tube lemma
Let us recall the thesis of the tube lemma. Assume, that and are topological spaces.
(TL) If is open (in product topology) and if is such that , then there exists an open neighbourhood of such that .
We wish to give a relation between (TL) and the the following thesis, concering closed projections:
(CP) The projection given by is a closed map.
The following theorem relates these two statements:
Theorem. (TL) is equivalent to (CP).
Proof. ,,” Let be a closed set and let be its open complement. We will show, that is closed, by showing that is open. So assume, that . Obviously
Therefore and by (TL) there exists open neighbourhood of such that . It easily follows, that and it is open, so (since was chosen arbitrary) is open.
,,” Let be an open subset such that for some . Let . Then is closed and by (CP) we have that is closed. Also and thus is an open neighbourhood of . It can be easily checked, that , 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 and are Hausdorff with compact, then both are true. On the other hand, for example for , where denotes reals with standard topology, they are both false. For example consider
Of course is closed, but 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 |