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I am trying to prove that all open balls in an ultrametric space are also closed. I found this proof (link below - page 3, #2) but I still don't understand it. Can someone explain their logic? What exactly do they mean?

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What you don't understand? – Makoto Kato Sep 26 '12 at 3:16
Do you know what an equivalence relation is? Do you know what a partition is? Do you know that equivalence relations partition underlying sets? – anon Sep 26 '12 at 3:24
"The requirement “x ∼ y if and only if d(x, y) < r” defines an equivalence relation on X whose classes are the open balls with radius r." I don't get what this is saying. What are classes of an equivalence relation? – user39794 Sep 26 '12 at 3:25
So you know what an equivalence relation is? If $X$ is a set and $\sim$ an equivalence relation, then the classes are of the form $$[x]:=\{y\in X: x\sim y\}.$$ Note that two classes are either the same or disjoint, and every element of $X$ is in an equivalence class. – anon Sep 26 '12 at 3:27
So what does ~ refer to in this ultrametric space? The distance? – user39794 Sep 26 '12 at 3:28
up vote 6 down vote accepted

You don’t need to use the machinery in that PDF to prove it.

By definition the ball $B(x,r^-)$ is open. Suppose that $y\in X\setminus B(x,r^-)$. Then $d(x,y)\ge r$, and I claim that $B(y,r^-)$ is an open ball around $y$ disjoint from $B(x,r^-)$. Assuming the claim for a moment, it follows that $X\setminus B(x,r^-)$ is open and hence that $B(x,r^-)$ is closed.

To prove the claim, suppose that $z\in B(y,r^-)\cap B(x,r^-)$. Then $$d(x,y)\le\max\{d(x,z),d(z,y)\}<r\;,$$ since $d$ is an ultrametric, contradicting the choice of $y$.

This shows that every open ball is also closed. Now we’ll show that the closed ball $B(x,r)$ is also open. Let $y\in B(x,r)$ be arbitrary; I claim that $B(y,r^-)\subseteq B(x,r)$, from which it follows immediately that $B(x,r)$ is open.

To prove the claim, let $z\in B(y,r^-)$, so that $d(z,y)<r$. We also have, $d(x,y)\le r$ by the choice of $y$. Thus $$d(z,x)\le\max\{d(z,y),d(y,x)\}\le r\;,$$ so $z\in B(x,r)$.

Added: However, it’s important that you learn to work with partitions and equivalence relations: they’re pretty nearly ubiquitous in mathematics, and they’re very handy tools. For the basics see the Wikipedia articles on equivalence relations, partitions, and equivalence classes and/or this page.

If $\langle X,d\rangle$ is an ultrametric space, and $r$ is any positive real number, we define a relation $\sim_r$ on $X$ as follows: $$x\sim_r y\quad\text{ iff }\quad d(x,y)\le r\;.$$

Clearly $x\sim_r x$ for every $x\in X$, since $d(x,x)=0\le r$, so $\sim_r$ is reflexive. The ultrametric $d$ is a symmetric function, so $x\sim_r y$ iff $y\sim_r x$, and $\sim_r$ is therefore symmetric. Finally, if $x\sim_r y$ and $y\sim_r z$, then $$d(x,z)\le\max\{d(x,y),d(y,z)\}\le r\;,$$ so $x\sim_r z$, and $\sim_r$ is transitive. By definition, then $\sim_r$ is an equivalence relation.

Now $B(x,r)=\{y\in X:d(x,y)\le r\}=\{y\in x:x\sim_r y\}$; that last set is by definition the $\sim_r$-equivalence class of $x$, so for each $x\in X$ the $\sim_r$-equivalence class of $x$ is simply the closed ball $B(x,r)$. It is a general fact about equivalence relations that equivalence classes are either disjoint or identical. In this case that means that for any $x,y\in X$, either $B(x,r)\cap B(y,r)=\varnothing$, or $B(x,r)=B(y,r)$. To put it in slightly different language, the equivalence classes of any equivalence relation form a partition of the underlying set: they divide it into parts in such a way that every point is in exactly one part.

Similarly, we can define

$$x\sim_{r^-} y\quad\text{ iff }\quad d(x,y)<r\;$$

and show that it too is an equivalence relation on $X$, and that its equivalence classes are the open balls $B(x,r^-)$ for $x\in X$. This is what’s being used in that PDF: since the balls $B(x,r^-)$ for $x\in X$ divide up $X$ in such a way that each point is in exactly one of them, the complement of any one of them is the union of all the others and is therefore open, being a union of open sets. This means that each one of them must be closed.

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Thanks Brian. I just deleted my other question. I was kind of upset because this user voted to close and downvote my two previous questions. In fact, they closed my previous question. Sometimes people are just too rough in here. – ILoveMath Nov 7 '14 at 10:36

Here's how to write the proof without using the terminology of equivalence relations.

Let $B(a,r)$ be an open ball of radius $r$ around the point $a$. Fix an arbitrary $x\in X\setminus B(a,r)$, and suppose the intersection $B(a,r)\cap B(x,r)$ is nonempty and say it contains a point $v$. Then

$$d(a,x)\le \min\{d(a,v),d(v,x)\}< r, \tag{$\star$}$$

the latter inequality because $d(a,v)<r$ (since $v\in B(a,r)$) and $d(v,x)<r$ (since $v\in B(x,r)$), but the above equation $(\star)$ is impossible because it implies $x\in B(a,r)$, contra hypothesis, so the original intersection $B(a,r)\cap B(x,r)$ is empty.

This allows us to write

$$X\setminus B(a,r)=\bigcup_{x\not\in B(a,r)}B(x,r),\tag{$\circ$}$$

because each point of the LHS is contained in a ball on the RHS, and each ball on the RHS is disjoint from the ball $B(a,r)$. Since $X\setminus B(a,r)$ is a union $(\circ)$ of open sets it is open, and hence we know that the open ball $B(a,r)$ is also a closed subset of $X$.

An equivalence relation $\sim$ is a generalization of equality on a set $X$, roughly speaking. Technically it's a subset of the cartesian product $X\times X$ in which the elements $(x,y)$ are written $x\sim y$. The usual axioms of equality are satisfied by an equivalence relation, namely symmetry, reflexivity, and of course transitivity (see the linked Wikipedia for details).

For each $x$, there is an equivalence class $[x]=\{y\in X: x\sim y\}$. Distinct equivalence classes are disjoint (contain no common element), and each element of $X$ is contained in some equivalence class, so the classes of $\sim$ partition the set $X$.

The pdf defines an equivalence relation on $X$ defined by $x\sim y$ precisely when $d(x,y)<r$; this is rarely an equivalence relation on a metric space, but it is one when our space is ultrametric, for by the argument above the balls $B(a,r)$ and $B(x,r)$ will be disjoint if $x$ is not in the equivalence class $[a]=B(a,r)$. (Here the equivalence classes are precisely the open balls, by definition.)

Essentially it's the same argument as above, but with the language of relations used in order to state that distinct open balls of the same radius are disjoint (when the center of one is outside of the other) and so they partition the space so that the complement of a ball $B$ of given radius is the union of all other balls of the same radius centered outside of $B$.

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