Vipul: Created page with '{{topospace property implication| stronger = metrizable space| weaker = perfectly normal space}} ==Statement== Any metrizable space, i.e., any space realized as the topolog…'

2009-10-26T16:18:15Z

Created page with '{{topospace property implication| stronger = metrizable space| weaker = perfectly normal space}} ==Statement== Any metrizable space, i.e., any space realized as the topolog…'

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{{topospace property implication|
stronger = metrizable space|
weaker = perfectly normal space}}

==Statement==

Any [[metrizable space]], i.e., any space realized as the topological space for a [[metric space]], is a [[perfectly normal space]] -- it is a [[normal space]] and every [[closed subset]] of it is a [[G-delta subset]] (it is a countable intersection of open subsets).

==Facts used==

# [[uses::Metrizable implies normal]]

==Proof==

'''Given''': A metric space <math>(X,d)</math>. with the topology arising from the metric.

'''To prove''': <math>X</math> is a perfectly normal space: <math>X</math> is a normal space and for every closed subset <math>A</math> of <math>X</math>, there is a countable collection of open subsets <math>U_n</math> of <math>X</math> such that <math>A</math> equals the intersection of the <math>U_n</math>s.

'''Proof''': By fact (1), <math>X</math> is a normal space, so we show the second part of the definition. For the closed subset <math>A</math>, define <math>U_n</math> as the set of all points <math>p \in X</math> such that there exists a point <math>a \in A</math> such that <math>d(p,a) < (1/n)</math>. Then:

# Each <math>U_n</math> is open: <math>U_n</math> is the union of the open balls of radius <math>1/n</math> about all the points of <math>A</math>. Hence, it is a union of open subsets, hence open.
# The intersection of the <math>U_n</math>s contains <math>A</math>.
# If <math>p</math> is not in <math>A</math>, there is some <math>U_n</math> such that <math>p \notin U_n</math>: Since <math>A</math> is closed, there exists <math>\epsilon</math> such that the ball of radius <math>\epsilon</math> about <math>p</math> does not intersect <math>A</math>. In other words, there is no point of <math>A</math> whose distance from <math>p</math> is less than <math>\epsilon</math>. Let <math>n</math> be a positive integer greater than <math>1/\epsilon</math>. Then, <math>U_n</math> does not contain <math>p</math>.

Together, (1), (2) and (3) complete the proof.

Metrizable implies perfectly normal - Revision history

Vipul: Created page with '{{topospace property implication| stronger = metrizable space| weaker = perfectly normal space}} ==Statement== Any metrizable space, i.e., any space realized as the topolog…'