Question on the Evenly Spaced Integer Topology

Question

Let $X$ be the set of integers with the topology generated by sets of the form $a + k\mathbb{Z} = \{a+k\lambda|\lambda \in \mathbb{Z}$}, where $a,k \in X$ and $k \neq 0$. The basis sets are sipmly the cosets of nonzero subgoups of the integers.

This is from the book "Counterexamples in Topology" by Steen/Seebach. It goes on to say that points in $X$ are closed but not open, and i'm having trouble making sense of that.

I would like somebody to explain to me the two different reasons why each point is closed, i.e. I'm wondering:

How is the complement of a point a union of basis elements?

How do i know that a set containing just a single point contains all its limit points?

Thanks everyone, I have a feeling this is going to be one of those "oh duh" questions once somebody points out to me the reasoning.

Answer 1

The set $\{0\}$ is closed because $\mathbb{Z}\setminus\{0\}$ is the union of the following open sets:

• $1+2\mathbb Z$;
• $2+4\mathbb Z$;
• $4+8\mathbb Z$;
• $\cdots$

A similar argument applies to any other integer.

In order to see that if $n\neq0$ then $n$ is not a limit point of $\{0\}$, just note that there are always integers $a$ and $k$, with $k\neq0$, such that $0\notin a+k\mathbb Z$ and that $n\in a+k\mathbb Z$; just take $a=n$ and $k=2n$, for instance. So, $n\in a+k\mathbb Z$ which is an open set to which $0$ does not bolong. Therefore, $n$ is not a limit point of $\{0\}$.

Answer 2

Given $x$, for every $y\ne x$ there exist $a$ and $k\ne 0$ such that $x\notin a+k\Bbb Z$ and $y\in a+k\Bbb Z$. Just pick $k=|x-y|+1$ and $a=y$. Hence the complement of $\{x\}$ is open.

---

Let $y$ be a limit point of a sequence living in $\{x\}$, i.e., of the constant sequence $x,x,\ldots$. Then every neighbourhood of $y$ contains some term of that sequence, i.e., contains $x$. By considering the neighbourhood $y+(|x-y|+1)\Bbb Z$, we find that $y=x$.

Answer 3

A family $B$ of subsets of a set $X$ is a base (basis) for a topology $T$ on $X$ iff

(i). $\cup B=X,$ and

(ii). if $b_1,b_2\in B$ and $s\in b_1\cap b_2$ then there exists $b_3\in B$ such that $s\in b_3\subset b_1\cap b_2.$

If $B$ is a base for a topology on $X$ then any non-empty $t\in T$ is the union of some members of $B$ so at least one $b\in B$ is a subset of $t.$ So if every member of $B$ is infinite then every non-empty $t\in T$ is infinite, and no $\{s\}$ can be open.

So in your Q, once we have proved that what we have is indeed a base for a topology, it follows that all non-empty open sets are infinite.

In your Q, if $a,a'\in X$ and $a\equiv a' \mod k$ for every $k\in \Bbb Z$ \ $\{0\}$ then $a=a'.$ So, given $a \in X,$ for each $a'\in X$ \ $\{a\}$ let $B(a')$ be a member of the base that contains $a'$ but not $a.$ Then $\cup_{a\ne a'\in X}B(a')$ is open , and its closed complement is $\{a\}.$

BTW. It is not hard to show that this topology is Hausdorff. Some deeper results imply that this topology on $\Bbb Z$ is actually homeomorphic to the usual topology on $\Bbb Q.$