Baire property is open subspace-closed - Revision history

Vipul: /* Proof */

2012-01-25T00:52:38Z

Proof

← Older revision		Revision as of 00:52, 25 January 2012
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	\| 6 \|\| The intersection <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B)</math> is dense in <math>X</math>. \|\| \|\| <math>X</math> is a Baire space. \|\| Step (5) \|\| Step-given direct		\| 6 \|\| The intersection <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B)</math> is dense in <math>X</math>. \|\| \|\| <math>X</math> is a Baire space. \|\| Step (5) \|\| Step-given direct
	\|-		\|-
	\| 7 \|\| <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B) = T \cup B</math> where <math>T = \bigcap_{n \in \mathbb{N}} U_n</math> \|\| \|\| \|\| \|\| \|\| pure set theory		\| 7 \|\| <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B) = T \cup B</math> where <math>T = \bigcap_{n \in \mathbb{N}} U_n</math> \|\| \|\| \|\| \|\| pure set theory
	\|-		\|-
	\| 8 \|\| <math>T \cup B</math> is dense in <math>X</math>. \|\| \|\| \|\| Steps (6), (7) \|\| Step-combination direct		\| 8 \|\| <math>T \cup B</math> is dense in <math>X</math>. \|\| \|\| \|\| Steps (6), (7) \|\| Step-combination direct
	\|-		\|-
	\| 9 \|\| <math>T</~~matH~~> is dense in <math>X</math>. In other words, for any nonempty open subset <math>S</math> of <math>A</math>, <math>S \cap T</math> is nonempty \|\| Fact (1) \|\| <math>A</math> is open in <math>X</math> \|\| Step (8) \|\| <toggledisplay>Since <math>S</math> is open in <math>A</math>, <math>S</math> is also open in <math>X</math> (by Fact (1)). By Step (8), <math>T \cup B</math> is dense in <math>X</math>, so <math>S \cap (T \cup B) = (S \cap T) \cup (S\cap B)</math> is nonempty. But since <math>S \subseteq A</math>, <math>S \cap B</math> is empty, so <math>S \cap T</math> must be nonempty, completing the proof.</toggledisplay>		\| 9 \|\| <math>T</math> is dense in <math>A</math>. In other words, for any nonempty open subset <math>S</math> of <math>A</math>, <math>S \cap T</math> is nonempty \|\| Fact (1) \|\| <math>A</math> is open in <math>X</math> \|\| Step (8) \|\| <toggledisplay>Since <math>S</math> is open in <math>A</math>, <math>S</math> is also open in <math>X</math> (by Fact (1)). By Step (8), <math>T \cup B</math> is dense in <math>X</math>, so <math>S \cap (T \cup B) = (S \cap T) \cup (S\cap B)</math> is nonempty. But since <math>S \subseteq A</math>, <math>S \cap B</math> is empty, so <math>S \cap T</math> must be nonempty, completing the proof.</toggledisplay>
	\|}		\|}

Vipul: /* Proof */

2012-01-25T00:49:43Z

Proof

← Older revision		Revision as of 00:49, 25 January 2012
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	{{further\|[[Subspace topology]]}}		{{further\|[[Subspace topology]]}}

			==Facts used==

			# [[uses::Open subset of open subspace is open]]
	==Proof==		==Proof==

	'''Given''': A Baire space <math>X</math>, an open subset <math>A</math>. A countable family of open dense subsets, <math>U_n, n \in \mathbb{N}</math> of <math>A</math>		'''Given''': A Baire space <math>X</math>, an open subset <math>A</math>. A countable family of open dense subsets, <math>U_n, n \in \mathbb{N}</math> of <math>A</math>

	'''To prove''': The intersection <math>\bigcap_{n \in \mathbb{N}} U_n</math> is dense in <math>A</math>		'''To prove''': The intersection <math>T = \bigcap_{n \in \mathbb{N}} U_n</math> is dense in <math>A</math>

	'''Proof''':		'''Proof''':
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	\| 1 \|\| Consider the set <math>B = X \setminus \overline{A}</math>, where <math>\overline{A}</math> is the closure of <math>A</math> in <math>X</math>. <math>B</math> is an '''open''' subset of <math>X</math>. \|\| \|\| \|\| \|\| the complement of a closed subset is open by definition.		\| 1 \|\| Consider the set <math>B = X \setminus \overline{A}</math>, where <math>\overline{A}</math> is the closure of <math>A</math> in <math>X</math>. <math>B</math> is an '''open''' subset of <math>X</math>. \|\| \|\| \|\| \|\| the complement of a closed subset is open by definition.
	\|-		\|-
	\| 2 \|\| Each <math>U_n</math> is open in <math>X</math>. \|\| \|\| <math>A</math> is open in <math>X</math><br>Each <math>U_n</math> is open in <math>A</math>. \|\| \|\|		\| 2 \|\| Each <math>U_n</math> is open in <math>X</math>. \|\| Fact (1) \|\| <math>A</math> is open in <math>X</math><br>Each <math>U_n</math> is open in <math>A</math>. \|\| \|\|
	\|-		\|-
	\| 3 \|\| Each <math>U_n \cup B</math> is open in <math>X</math> with <math>B</math> defined in Step (1). \|\| \|\| \|\| Steps (1), (2) \|\| Step-combination, and the observation that a union of open subsets is open		\| 3 \|\| Each <math>U_n \cup B</math> is open in <math>X</math> with <math>B</math> defined in Step (1). \|\| \|\| \|\| Steps (1), (2) \|\| Step-combination, and the observation that a union of open subsets is open
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	\|-		\|-
	\| 6 \|\| The intersection <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B)</math> is dense in <math>X</math>. \|\| \|\| <math>X</math> is a Baire space. \|\| Step (5) \|\| Step-given direct		\| 6 \|\| The intersection <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B)</math> is dense in <math>X</math>. \|\| \|\| <math>X</math> is a Baire space. \|\| Step (5) \|\| Step-given direct
			\|-
			\| 7 \|\| <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B) = T \cup B</math> where <math>T = \bigcap_{n \in \mathbb{N}} U_n</math> \|\| \|\| \|\| \|\| \|\| pure set theory
			\|-
			\| 8 \|\| <math>T \cup B</math> is dense in <math>X</math>. \|\| \|\| \|\| Steps (6), (7) \|\| Step-combination direct
			\|-
			\| 9 \|\| <math>T</matH> is dense in <math>X</math>. In other words, for any nonempty open subset <math>S</math> of <math>A</math>, <math>S \cap T</math> is nonempty \|\| Fact (1) \|\| <math>A</math> is open in <math>X</math> \|\| Step (8) \|\| <toggledisplay>Since <math>S</math> is open in <math>A</math>, <math>S</math> is also open in <math>X</math> (by Fact (1)). By Step (8), <math>T \cup B</math> is dense in <math>X</math>, so <math>S \cap (T \cup B) = (S \cap T) \cup (S\cap B)</math> is nonempty. But since <math>S \subseteq A</math>, <math>S \cap B</math> is empty, so <math>S \cap T</math> must be nonempty, completing the proof.</toggledisplay>
	\|}		\|}

	~~The rest of the proof needs to be converted to tabular form too!~~

	so their intersection is dense in <math>X</math>. This intersection is of the form <math>B \cup T</math>, where <math>T</math> is an open subset of <math>A</math>. We now need to argue that <math>T</math> is a dense subset of <math>A</math>.

	Let <math>S</math> be a nonempty open subset of <math>A</math>; we need to show that <math>S \cap T</math> is nonempty. Since <math>S</math> is open in <math>A</math>, <math>S</math> is also open in <math>X</math>, so <math>S \cap (T \cup B)</math> is nonempty, because <math>T \cup B</math> is dense in <math>X</math>. But since <math>S \subset A</math>, <math>S \cap B</math> is empty, so <math>S \cap T</math> must be nonempty, completing the proof.

	==References==		==References==

Vipul: /* Proof */

2012-01-25T00:20:15Z

Proof

← Older revision		Revision as of 00:20, 25 January 2012
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	\| 5 \|\| Consider the collection of subsets <math>U_n \cup B</math>. This is a countable collection of open dense subsets of <math>X</math> \|\| \|\| \|\| Steps (3), (4) \|\| Step-combination direct		\| 5 \|\| Consider the collection of subsets <math>U_n \cup B</math>. This is a countable collection of open dense subsets of <math>X</math> \|\| \|\| \|\| Steps (3), (4) \|\| Step-combination direct
	\|-		\|-
	\| 6 \|\| The intersection <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B)</math> is dense in <math>X</math>> \|\| \|\| <math>X</math> is a Baire space. \|\| Step (5) \|\| Step-given direct		\| 6 \|\| The intersection <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B)</math> is dense in <math>X</math>. \|\| \|\| <math>X</math> is a Baire space. \|\| Step (5) \|\| Step-given direct
	\|}		\|}

Vipul: /* Proof */

2012-01-25T00:19:47Z

Proof

← Older revision		Revision as of 00:19, 25 January 2012
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	==Proof==		==Proof==

	'''Given''': A Baire space <math>X</math>, an open subset <math>A</math>		'''Given''': A Baire space <math>X</math>, an open subset <math>A</math>. A countable family of open dense subsets, <math>U_n, n \in \mathbb{N}</math> of <math>A</math>

	'''To prove''': An intersection ~~of a countable family of open dense subsets,~~ <math>~~U_n,~~ n \in \mathbb{N}~~</math> of <math>A~~</math> is dense in <math>A</math>		'''To prove''': The intersection <math>\bigcap_{n \in \mathbb{N}} U_n</math> is dense in <math>A</math>

	'''Proof''': ~~First, consider the set <math>B = X \setminus \overline{A}</math>, where <math>\overline{A}</math> is the closure of <math>A</math> in <math>X</math>. Clearly <math>B</math> is open in <math>X</math>.~~		'''Proof''':

	~~Since~~ <math>A</math> is open in <math>X</math>, and ~~each~~ <math>U_n</math> is open in <math>A</math>, ~~each~~ <math>U_n</math> is open in <math>X</math>. Hence, ~~each~~ <math>U_n \cup B</math> is open in <math>X</math>.		{\| class="sortable" border="1"
			! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
			\|-
			\| 1 \|\| Consider the set <math>B = X \setminus \overline{A}</math>, where <math>\overline{A}</math> is the closure of <math>A</math> in <math>X</math>. <math>B</math> is an '''open''' subset of <math>X</math>. \|\| \|\| \|\| \|\| the complement of a closed subset is open by definition.
			\|-
			\| 2 \|\| Each <math>U_n</math> is open in <math>X</math>. \|\| \|\| <math>A</math> is open in <math>X</math><br>Each <math>U_n</math> is open in <math>A</math>. \|\| \|\|
			\|-
			\| 3 \|\| Each <math>U_n \cup B</math> is open in <math>X</math> with <math>B</math> defined in Step (1). \|\| \|\| \|\| Steps (1), (2) \|\| Step-combination, and the observation that a union of open subsets is open
			\|-
			\| 4 \|\| Each <math>U_n \cup B</math> is dense in <math>X</math>. In other words, for any open subset <math>V</math> of <math>X</math>, the intersection <math>(U_n \cup B) \cap V</math> is nonempty. \|\| \|\| \|\| \|\| <toggledisplay>Consider any open subset <math>V</math> of <math>X</math>. If the open subset intersects <math>B</math>, we are done. Otherwise the open subset is in <math>\overline{A}</math>. Hence, by definition, it intersects <math>A</math> in a nonempty open subset, say <math>W</math>. Then <math>W</math> is an open subset of <math>A</math>. Since <math>U_n</math> is dense in <math>A</math>, it intersects <math>W</math> nontrivially, so it intersects <math>V</math> nontrivially, completing the proof of density.</toggledisplay>
			\|-
			\| 5 \|\| Consider the collection of subsets <math>U_n \cup B</math>. This is a countable collection of open dense subsets of <math>X</math> \|\| \|\| \|\| Steps (3), (4) \|\| Step-combination direct
			\|-
			\| 6 \|\| The intersection <math>\bigcap_{n \in \mathbb{N}} (U_n \cup B)</math> is dense in <math>X</math>> \|\| \|\| <math>X</math> is a Baire space. \|\| Step (5) \|\| Step-given direct
			\|}

	Next, we want to argue that each <math>U_n \cup B</math> is dense in <math>X</math>. For this, consider any open subset <math>V</math> of <math>X</math>. If the open subset intersects <math>B</math>, we are done. Otherwise the open subset is in <math>\overline{A}</math>. Hence, by definition, it intersects <math>A</math> in a nonempty open subset, say <math>W</math>. Then <math>W</math> is an open subset of ~~<math>A</math>. Since <math>U_n</math> is dense in <math>A</math>, it intersects <math>W</math> nontrivially, completing~~ the proof ~~of density.~~		The rest of the proof needs to be converted to tabular form too!

	~~Thus, consider the collection of subsets <math>U_n \cup B</math>. This is a countable collection of open dense subsets of <math>X</math>,~~ so their intersection is dense in <math>X</math>. This intersection is of the form <math>B \cup T</math>, where <math>T</math> is an open subset of <math>A</math>. We now need to argue that <math>T</math> is a dense subset of <math>A</math>.		so their intersection is dense in <math>X</math>. This intersection is of the form <math>B \cup T</math>, where <math>T</math> is an open subset of <math>A</math>. We now need to argue that <math>T</math> is a dense subset of <math>A</math>.

	Let <math>S</math> be a nonempty open subset of <math>A</math>; we need to show that <math>S \cap T</math> is nonempty. Since <math>S</math> is open in <math>A</math>, <math>S</math> is also open in <math>X</math>, so <math>S \cap (T \cup B)</math> is nonempty, because <math>T \cup B</math> is dense in <math>X</math>. But since <math>S \subset A</math>, <math>S \cap B</math> is empty, so <math>S \cap T</math> must be nonempty, completing the proof.		Let <math>S</math> be a nonempty open subset of <math>A</math>; we need to show that <math>S \cap T</math> is nonempty. Since <math>S</math> is open in <math>A</math>, <math>S</math> is also open in <math>X</math>, so <math>S \cap (T \cup B)</math> is nonempty, because <math>T \cup B</math> is dense in <math>X</math>. But since <math>S \subset A</math>, <math>S \cap B</math> is empty, so <math>S \cap T</math> must be nonempty, completing the proof.

Vipul at 00:06, 25 January 2012

2012-01-25T00:06:43Z

← Older revision		Revision as of 00:06, 25 January 2012
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	{{topospace metaproperty satisfaction}}		{{topospace metaproperty satisfaction\|
			property = Baire space\|
			metaproperty = open subspace-closed property of topological spaces}}

	==Statement==		==Statement==

Vipul at 20:08, 21 July 2008

2008-07-21T20:08:57Z

← Older revision		Revision as of 20:08, 21 July 2008
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	Every [[open subset]] of a [[Baire space]] is itself a [[Baire space]], under the [[subspace topology]].		Every [[open subset]] of a [[Baire space]] is itself a [[Baire space]], under the [[subspace topology]].

			==Definitions used==

			===Baire space===

			{{further\|[[Baire space]]}}

			A [[topological space]] is termed a '''Baire space''' if an intersection of countably many open dense subsets is dense.

			===Subspace topology===

			{{further\|[[Subspace topology]]}}

			==Proof==

			'''Given''': A Baire space <math>X</math>, an open subset <math>A</math>

			'''To prove''': An intersection of a countable family of open dense subsets, <math>U_n, n \in \mathbb{N}</math> of <math>A</math> is dense in <math>A</math>

			'''Proof''': First, consider the set <math>B = X \setminus \overline{A}</math>, where <math>\overline{A}</math> is the closure of <math>A</math> in <math>X</math>. Clearly <math>B</math> is open in <math>X</math>.

			Since <math>A</math> is open in <math>X</math>, and each <math>U_n</math> is open in <math>A</math>, each <math>U_n</math> is open in <math>X</math>. Hence, each <math>U_n \cup B</math> is open in <math>X</math>.

			Next, we want to argue that each <math>U_n \cup B</math> is dense in <math>X</math>. For this, consider any open subset <math>V</math> of <math>X</math>. If the open subset intersects <math>B</math>, we are done. Otherwise the open subset is in <math>\overline{A}</math>. Hence, by definition, it intersects <math>A</math> in a nonempty open subset, say <math>W</math>. Then <math>W</math> is an open subset of <math>A</math>. Since <math>U_n</math> is dense in <math>A</math>, it intersects <math>W</math> nontrivially, completing the proof of density.

			Thus, consider the collection of subsets <math>U_n \cup B</math>. This is a countable collection of open dense subsets of <math>X</math>, so their intersection is dense in <math>X</math>. This intersection is of the form <math>B \cup T</math>, where <math>T</math> is an open subset of <math>A</math>. We now need to argue that <math>T</math> is a dense subset of <math>A</math>.

			Let <math>S</math> be a nonempty open subset of <math>A</math>; we need to show that <math>S \cap T</math> is nonempty. Since <math>S</math> is open in <math>A</math>, <math>S</math> is also open in <math>X</math>, so <math>S \cap (T \cup B)</math> is nonempty, because <math>T \cup B</math> is dense in <math>X</math>. But since <math>S \subset A</math>, <math>S \cap B</math> is empty, so <math>S \cap T</math> must be nonempty, completing the proof.

	==References==		==References==

Vipul: New page: {{topospace metaproperty satisfaction}} ==Statement== ===Verbal statement=== Every open subset of a Baire space is itself a Baire space, under the subspace topology. ==...

2008-07-20T22:07:47Z

New page: {{topospace metaproperty satisfaction}} ==Statement== ===Verbal statement=== Every open subset of a Baire space is itself a Baire space, under the subspace topology. ==...

New page

{{topospace metaproperty satisfaction}}

==Statement==

===Verbal statement===

Every [[open subset]] of a [[Baire space]] is itself a [[Baire space]], under the [[subspace topology]].

==References==

===Textbook references===
* {{booklink-proved|Munkres}}, Page 297, Lemma 48.4, Chapter 4, Section 48