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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

1 Answers

One Answer

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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