First-countable implies compactly generated - Revision history

Vipul: /* Proof that this collection works */

2009-10-26T20:16:18Z

Proof that this collection works

← Older revision		Revision as of 20:16, 26 October 2009
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	===Proof that this collection works===		===Proof that this collection works===

	'''Given''': A subset <math>V</math> of <math>X</math>, such that <math>V \cap K</math> is open in <math>K</math> for every <math>K</math> of the above form.		'''Given''': A subset <math>V</math> of <math>X</math> such that <math>V \cap K</math> is open in <math>K</math> for every <math>K</math> of the above form.

	'''To prove''': <math>V</math> is open in <math>X</math>		'''To prove''': <math>V</math> is open in <math>X</math>
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	'''Proof''': We want to show that for every <math>x \in V</math>, there exists an open subset of <math>X</math> containing <math>x</math> and inside <math>V</math>.		'''Proof''': We want to show that for every <math>x \in V</math>, there exists an open subset of <math>X</math> containing <math>x</math> and inside <math>V</math>.

	Since <math>X</math> is first-countable, we can find, for any point <math>x \in X</math>, a descending chain <math>U_n</math> of open subsets containing <math>x</math>, such that every open set containing <math>x</math> contains one of the <math>U_n</math>s. ~~{{fillin}}~~		Since <math>X</math> is first-countable, we can find, for any point <math>x \in X</math>, a descending chain <math>U_n</math> of open subsets containing <math>x</math>, such that every open set containing <math>x</math> contains one of the <math>U_n</math>s. It thus suffices to show that one of the <math>U_n</math>s is contained in <math>V</math>.

			Suppose not. Then, none of the <math>U_n</math>s is contained in <math>V</math>. Pick <math>a_n \in U_n \setminus V</math> for each <math>n</math>, and consider the sequence of <math>a_n</math>s. Any open subset containing <math>x</math> contains one of the <math>U_n</math>s, so it contains all but finitely many of the <math>a_n</math>s. Thus, the <math>a_n</math>s converge to <math>x</math> Consider the compact set <math>K</math> formed by the <math>a_n</math>s and <math>x</math>. By assumption, <math>V \cap K</math> is open in <math>K</math>. But this would force <math>V \cap K</math> to contain one or more of the <math>a_n</math>s, a contradiction. This completes the proof.

Vipul: /* Proof that this collection works */

2009-10-26T19:52:07Z

Proof that this collection works

← Older revision		Revision as of 19:52, 26 October 2009
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	'''To prove''': <math>V</math> is open in <math>X</math>		'''To prove''': <math>V</math> is open in <math>X</math>

	'''Proof''': ~~Suppose <math>V</math> is not open.~~ We want to show that for every <math>x \in V</math>, there exists an open subset of <math>X</math> containing <math>x</math> and inside <math>V</math>.		'''Proof''': We want to show that for every <math>x \in V</math>, there exists an open subset of <math>X</math> containing <math>x</math> and inside <math>V</math>.

	Since <math>X</math> is first-countable, we can find, for any point <math>x \in X</math>, a descending chain <math>U_n</math> of open subsets containing <math>x</math>, such that every open set containing <math>x</math> contains one of the <math>U_n</math>s.		Since <math>X</math> is first-countable, we can find, for any point <math>x \in X</math>, a descending chain <math>U_n</math> of open subsets containing <math>x</math>, such that every open set containing <math>x</math> contains one of the <math>U_n</math>s. {{fillin}}

Vipul at 11:15, 8 August 2008

2008-08-08T11:15:45Z

← Older revision		Revision as of 11:15, 8 August 2008
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	{{topospace property implication}}		{{topospace property implication\|
			stronger = first-countable space\|
			weaker = compactly generated space}}

	==Statement==		==Statement==

Vipul at 00:25, 22 July 2008

2008-07-22T00:25:40Z

← Older revision		Revision as of 00:25, 22 July 2008
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	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>.		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		Let's see why <math>K</math> is compact. First, observe that by construction, any open set of <math>K</math> containing <math>x</math> must contain all but finitely many <math>x_n</math>. Hence, if we have an open cover of <math>K</math>, the member of the open cover containing <math>x</math> must contain all but finitely many of the <math>x_n</math>s. Picking one member of the open cover for each <math>x_n</math>, we obtain a finite subcover.

			===Proof that this collection works===

			'''Given''': A subset <math>V</math> of <math>X</math>, such that <math>V \cap K</math> is open in <math>K</math> for every <math>K</math> of the above form.

			'''To prove''': <math>V</math> is open in <math>X</math>

			'''Proof''': Suppose <math>V</math> is not open. We want to show that for every <math>x \in V</math>, there exists an open subset of <math>X</math> containing <math>x</math> and inside <math>V</math>.

			Since <math>X</math> is first-countable, we can find, for any point <math>x \in X</math>, a descending chain <math>U_n</math> of open subsets containing <math>x</math>, such that every open set containing <math>x</math> contains one of the <math>U_n</math>s.

Vipul: /* Proof */

2008-07-21T21:52:18Z

Proof

← Older revision		Revision as of 21:52, 21 July 2008
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	==Proof==		==Proof==

	{{~~fillin~~}}		'''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

Vipul: 1 revision

2008-05-11T19:44:37Z

1 revision

← Older revision	Revision as of 19:44, 11 May 2008
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Vipul: New page: {{topospace property implication}} ==Statement== ===Property-theoretic statement=== The property of topological spaces of being first-countable is stronger...

2008-02-02T01:19:27Z

New page: {{topospace property implication}} ==Statement== ===Property-theoretic statement=== The property of topological spaces of being first-countable is stronger...

New page

{{topospace property implication}}

==Statement==

===Property-theoretic statement===

The [[property of topological spaces]] of being [[first-countable space|first-countable]] is stronger than the property of being [[compactly generated space|compactly generated]].

===Verbal statement===

Any [[first-countable space]] is [[compactly generated space|compactly generated]].

==Proof==

{{fillin}}