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Let X:=R^n with the Euclidean Topology. Is X first countable? Find a nested basis

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Homework Statement


Let X:=ℝn with the Euclidean Topology. Is X first countable? Find a nested neighborhood basis for X at 5.


Homework Equations


If X is a topological space and p[itex]\in[/itex]X, a collection [itex]B[/itex]p of neighborhoods of p is called a neighborhood basis for X at p if every neighborhood of p contains some B[itex]\in[/itex][itex]B[/itex]p.

We say X is first countable if there exists a countable neighborhood basis at each point.


The Attempt at a Solution


I say yes.
Let p[itex]\in[/itex]X, the set of open balls Br(p) for r being rational forms a neighborhood basis at p. (That is, for all neighborhoods U of p, there is a Br(p)[itex]\subseteq[/itex]U)
Since p was arbitrary and this [itex]B[/itex]p is countable (since rationals are countable), X is first countable.

As well, we can just let the nest interval be defined as: B(1/2)i(5) for i being a natural number. Thus, B(1/2)i+1(5)<B(1/2)i(5).

I am struggling a bit at this level of proof honestly, and I'm trying to stay afloat. Thank you!
 

Answers and Replies

  • #2
jbunniii
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Your proof looks fine. Just a few nitpicking details: first, [itex]r[/itex] needs to be rational AND positive. Second, although it's pretty obvious, you might say a few words about why you can find an [itex]r[/itex] such that [itex]B_r(p) \subseteq U[/itex].

Your nested neighborhood basis at 5 is fine. Just one minor detail: instead of <, you want [itex]\subset[/itex].
 
  • #3
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Thank you!

Wouldn't it just be based on the density of the rationals?
Also, thank you for catching my typo :)
 
  • #4
dextercioby
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Yes, that is correct.
 
  • #5
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Thanks again! This site is amazing!
 

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