# Section 2 E, problem (b) Kelley

I think I finally managed to prove this exercise from Kelley’s general topology on page 77.
I would appreciate feedback on it.

Let $$X$$ be the set of all pairs of $$Bbb N_0$$ with the topology described as follows:
For each point $$(m,n)$$ other than $$(0,0)$$ the set $${(m,n)}$$ is open.
A set $$U$$ is a neighbourhood of $$(0,0)$$ iff for all except a finite number of integers $$m$$ the set $${n:(m,n)notin U}$$ is finite.
(Visualizing $$X$$ in the Euclidean plane, a neighbourhood of $$(0,0)$$ contains all but a finite number of the members of all but a finite number of columns.)

(b) Each point of $$X$$ is the intersection of a countable family of closed neghbourhoods

Proof
Since any point other than $$(0,0)$$ is open, complement of the union of all open points is $$(0,0)$$. Hence it is closed.
So let $$(a,b)$$ be any point in $$X$$, let $$mathcal U$$ be a countable family of neighbourhoods of $$(0,0)$$ s.t:

• Every member of $$mathcal U$$ contains $$(a,b)$$;
• each member excludes finite amount of columns and finite points from column of $$(a,b)$$ which no other member excludes.

Then $$cap^infty {U:, Uinmathcal U} = (a,b).$$

Mathematics Asked on November 21, 2021

It has a few problems. I think that you have the right idea for showing that $${langle 0,0rangle}$$ is the intersection of a countable family of closed nbhds of $$langle 0,0rangle$$, but you’ve not stated it very clearly. For convenience let $$D=Xsetminus{langle 0,0rangle}$$. For each $$langle m,nranglein D$$ let $$U(m,n)=Xsetminus{langle m,nrangle};;$$ $$U(m,n)$$ is a nbhd of $$langle 0,0rangle$$, and since the point $${langle m,nrangle}$$ is open, $$U(m,n)$$ is closed set. Clearly the family $${U(m,n):langle m,nranglein D}$$ is countable, and

begin{align*} &bigcapbig{U(m,n):langle m,nranglein Xsetminus{langle 0,0rangle}big}\ &quad=bigcapbig{Xsetminus{langle m,nrangle}:langle m,nranglein Xsetminus{langle 0,0rangle}big}\ &quad=Xsetminusbigcupbig{{langle m,nrangle}:langle m,nranglein Xsetminus{langle 0,0rangle}big}\ &quad=Xsetminus(Xsetminus{langle 0,0rangle})\ &quad={langle 0,0rangle};, end{align*}

so $${langle 0,0rangle}$$ is the intersection of a countable family of closed nbhds of $$langle 0,0rangle$$.

Your argument for the other points, however, definitely does not work: the members of your family $$mathscr{U}$$ are all nbhds of $$langle 0,0rangle$$, so their intersection contains $$langle 0,0rangle$$ and therefore cannot be $${langle a,brangle}$$.

Note that in any case the intersection is a set, not a point. If you had a family $$mathscr{U}$$ that actually worked, its intersection would still not be $$langle a,brangle$$: it would be $${langle a,brangle}$$, the set whose only member is the point $$langle a,brangle$$.

In fact if $$langle m,nranglein D$$, all you need to do is let $$V={langle m,nrangle}$$ and let $$mathscr{V}={V}$$. Certainly $$V$$ is a nbhd of $$langle m,nrangle$$, $$mathscr{V}$$ is countable, and $$bigcapmathscr{V}={langle m,nrangle}$$, so it only remains to verify that $$V$$ is a closed nbhd of $$langle m,nrangle$$. But $$V=Xsetminus U(m,n)$$, and we already observed that $$U(m,n)$$ is open nbhd, so $$V$$ is closed, and we’re done.

Answered by Brian M. Scott on November 21, 2021

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