Is the Euclidean Topology of ℝn First Countable?

[Hodgey8806]

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$\in$X, a collection $B$_p of neighborhoods of p is called a neighborhood basis for X at p if every neighborhood of p contains some B$\in$$B$_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$\in$X, the set of open balls B_r(p) for r being rational forms a neighborhood basis at p. (That is, for all neighborhoods U of p, there is a B_r(p)$\subseteq$U)
Since p was arbitrary and this $B$_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!

 

[jbunniii]

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

Your nested neighborhood basis at 5 is fine. Just one minor detail: instead of <, you want $\subset$.

 

[Hodgey8806]

Thank you!

Wouldn't it just be based on the density of the rationals?
Also, thank you for catching my typo :)

 

[dextercioby]

Yes, that is correct.

 

[Hodgey8806]

Thanks again! This site is amazing!