The Unapologetic Mathematician

Separation Properties

There’s a whole list of properties of topological spaces that we may want to refer to called the separation axioms. Even when two points are distinct elements of the underlying set of a topological space, we may not be able to tell them apart with topological techniques. Points are separated if we can tell them apart in some way using the topology. Today we’ll discuss various properties of separation, and tomorrow we’ll list some of the more useful separation axioms we can ask that a space satisfy.

First, and weakest, we say that points $x$ and $y$ in a topological space $X$ are “topologically distinguishable” if they don’t have the same collection of neighborhoods — if $\mathcal{N}(x)\neq\mathcal{N}(y)$. Now maybe one of the collections of neighborhoods strictly contains the other: $\mathcal{N}(x)\subseteq\mathcal{N}(y)$. In this case, every neighborhood of $x$ is a neighborhood of $y$. a forteriori it contains a neighborhood of $y$, and thus contains $y$ itself. Thus the point $x$ is in the closure of the set $\{y\}$. This is really close. The points are topologically distinguishable, but still a bit too close for comfort. So we define points to be “separated” if each has a neighborhood the other one doesn’t, or equivalently if neither is in the closure of the other. We can extend this to subsets larger than just points. We say that two subsets $A$ and $B$ are separated if neither one touches the closure of the other. That is, $A\cap\bar{B}=\varnothing$ and $\bar{A}\cap B=\varnothing$.

We can go on and give stronger conditions, saying that two sets are “separated by neighborhoods” if they have disjoint neighborhoods. That is, there are neighborhoods $U$ and $V$ of $A$ and $B$, respectively, and $U\cap V=\varnothing$. Being a neighborhood here means that $U$ contains some open set $S$ which contains $A$ and $V$ contains some open set $T$ which contains $B$, and so the closure of $A$ is contained in the open set, and thus in $U$. Similarly, the closure of $B$ must be contained in $V$.. We see that the closure of $B$ is contained in the complement of $S$, and similarly the closure of $A$ is in the complement of $T$, so neither $A$ nor $B$ can touch the other’s closure. Stronger still is being “separated by closed neighborhoods”, which asks that $U$ and $V$ be disjoint closed neighborhoods. These keep $A$ and $B$ even further apart, since these neighborhoods themselves can’t touch each other’s closures.

The next step up is that sets be “separated by a function” if there is a continuous function $f:X\rightarrow\mathbb{R}$ so that for every point $a\in A$ we have $f(a)=0$, and for every point $b\in B$ we have $f(b)=1$. In this case we can take the closed interval $\left[-1,\frac{1}{3}\right]$ whose preimage must be a closed neighborhood of $A$ by continuity. Similarly we can take the closed interval $\left[\frac{2}{3},2\right]$ whose preimage is a closed neighborhood of $B$. Since these preimages can’t touch each other, we have separated $A$ and $B$ by closed neighborhoods. Stronger still is that $A$ and $B$ are “precisely separated by a function”, which adds the requirement that only points from $A$ go to ${0}$ and only points from $B$ go to $1$.

This list of separation

January 10, 2008 - Posted by | Point-Set Topology, Topology

1. “Being a neighborhood here means that U contains some open set which contains A, and so che closure of A is contained in the open set, and thus in U”

I think I’m not getting this…

Comment by edriv | January 10, 2008 | Reply

2. Okay.. remember how we defined a neighborhood of a point? It doesn’t just contain the point, it contains an open set containing the point. So here, we start with the set $A$ and say that $U$ is a neighborhood of $A$ if there is an open set $S$ with $A\subseteq S\subseteq U$.

The open set $S$ provides the “wiggle room” we need, and $U$ has at least as much as $S$ does.

Comment by John Armstrong | January 10, 2008 | Reply

3. The problem is the closure, maybe I’m misreading, but haven’t you written that if A is a set and O (open set) contains A, then O contains the closure of A?

Comment by edriv | January 10, 2008 | Reply

4. Oh, gr.. I went too fast there and wrote down the wrong line of reasoning. Let me fix that reasoning in the original.

Comment by John Armstrong | January 10, 2008 | Reply

5. [...] Now that we have some vocabulary about separation properties down we can talk about properties of spaces as a whole, called the separation [...]

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