This is (part of) Exercise 7 from Section 2.2 on page 22 of Topology and Groupoids, by Brown. I would appreciate any feedback regarding the quality of my proof.
Exercise:
Let $X = \mathbb{Z}$ and let $p$ be a fixed integer. A set $N
\subseteq \mathbb{Z}$ is a p-adic neighborhood of $n \in
\mathbb{Z}$ if $N$ contains the integers $n + mp^r$ for some $r$ and
all $m = 0, \pm 1, \pm 2, \dots$ (so that in a given neighbourhood $r$
is fixed but $m$ varies). Prove that the $p$-adic neighborhoods form a
neighborhood topology on $\mathbb{Z}$, the $p$-adic topology.
Is this topology the same as the order topology? The discrete
topology? The indiscrete topology?
More information:
The neighborhood axioms I am using are as follows:
- If $N$ is a neighborhood of $x$, then $x \in N$.
- If $N$ is a subset of $X$ containing a neighbourhood of $x$, then $N$ is a neighbourhood of $x$.
- The intersection of two neighbourhoods of $x$ is again a neighbourhood of $x$.
- Any neighbourhood $N$ of $x$ contains a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$.
My attempt:
First axiom:
If $N$ is a neighborhood of $x$, then clearly $x \in N$ because we can let $m = 0$ in the expression $x + mp^r$.
Second axiom:
If $M$ is a neighborhood of $x$, and $M \subseteq N$, then clearly $N$ contains all the numbers $x + mp^r$ that $M$ contains, so $N$ must be a neighborhood of $x$ as well.
Third axiom:
If $M$ and $N$ are neighborhoods of $x$, then all $x + mp^{r_M} \in M$, and all $x + mp^{r_N} \in N$ where in each case $m \in \mathbb{Z}$ and $p, r_M$ and $r_N$ are all fixed integers. If $r_M = r_N$ then call the common exponent simply $r$, and both sets contain all $x + mp^r$ so their intersection must still be a neighborhood of $x$.
If $r_M \neq r_N$, then assume without loss of generality that $r_M < r_N$. Then we may write $r_N = r_M + k$, where $k \in \mathbb{N}$, so that any number $x + mp^{r_N} \in N$ may be written as $x + mp^k p^{r_M}$ which is clearly an element of $M$. Therefore, all the $x + mp^{r_N} \in N$ are also in $M$, so the intersection $M \cap N$ is still a neighborhood of $x$.
Fourth axiom:
Assume that $N$ is a neighborhood of $x$. That means $N$ contains all numbers $x + mp^r$ for $m \in \mathbb{Z}$, fixed $p$ and fixed $r$. Now consider $M = \{ x + mp^r \colon m \in \mathbb{Z} \}$. In other words $M$ is only those numbers $x + mp^r$. Then $M$ is also a neighborhood of $x$ and $M \subseteq N$. We need to show that $N$ is a neighborhood of some $x + m_0 p^r \in M$. In other words, $N$ must contain all numbers $(x + m_0 p^r) + mp^r$. This is true because $(x + m_0 p^r) + mp^r = x + (m_0 + m)p^r$ which is clearly an element of $N$ because $m_0 + m$ is just another integer.
Comparisons to the order, discrete and indiscrete topologies:
The $p$-adic topology on $\mathbb{Z}$ is distinct from the order topology on $\mathbb{Z}$ (which is equivalent to the discrete topology on $\mathbb{Z}$) because the order topology allows for finite neighborhoods, while the $p$-adic topology does not.
The $p$-adic topology on $\mathbb{Z}$ is also distinct from the indiscrete topology because the indiscrete topology allows only $\mathbb{Z}$ itself to be a neighborhood of any point, whereas the $p$-adic topology also allows smaller sets to be neighborhoods of a point.
Best Answer
This is fine. My only comment is that there’s no reason to split the argument for the third axiom into cases. Just let $r=\max\{r_M,r_N\}$, and observe that $$\{n+mp^r:m\in\Bbb Z\}\subseteq\{n+mp^s:m\in\Bbb Z\}$$ whenever $r\ge s$, since $n+mp^r=n+mp^{r-s}p^s$.