First-countable implies compactly generated: Difference between revisions
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==Proof== | ==Proof== | ||
{{ | '''Given''': A first-countable space <math>X</math> | ||
'''To prove''': There exists a collection of compact subsets <math>\{ K_i \}_{i \in I}</math> of <math>X</math>, such that <math>U \subset X</math> is open if and only if <math>U \cap K_i</math> is open in <math>K_i</math> for every <math>i</math> | |||
===Construction of the collection of compact subsets=== | |||
We consider compact subsets of the following form: take a sequence <math>x_n</math> that converges to a point <math>x</math> (note that a sequence may converge to more than one point, we just make sure there is at least one point of convergence). Now define the compact set <math>K</math> corresponding to this sequence as <math> \{ x_n \}_{n \in \mathbb{N}} \cup \{ x \}</math>. | |||
Let's first see why <math>K</math> is compact. In |
Revision as of 21:52, 21 July 2008
This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property must also satisfy the second topological space property
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Statement
Property-theoretic statement
The property of topological spaces of being first-countable is stronger than the property of being compactly generated.
Verbal statement
Any first-countable space is compactly generated.
Proof
Given: A first-countable space
To prove: There exists a collection of compact subsets of , such that is open if and only if is open in for every
Construction of the collection of compact subsets
We consider compact subsets of the following form: take a sequence that converges to a point (note that a sequence may converge to more than one point, we just make sure there is at least one point of convergence). Now define the compact set corresponding to this sequence as .
Let's first see why is compact. In