Locally Confused

Adding “companion points”

last edited around 2019-02-09

Let $X$ be a topological space, $S\subseteq X$. Define $\DeclareMathOperator{\comp}{Comp}\comp(X, p)$, to be the set $X\sqcup {p}$ equipped with the topology $$ \{U\cup \{p\} \mid U \ \text{Open in}\ X\} \cup \emptyset. $$ It is straightforward to check that this indeed defines a topology. Our new point $p$ drastically introduces new “convergence” behavior: Because $\{p\}$ is open, the only way for a sequence to converge against $p$ is to be constantly $p$ at some $N\in \mathbb N$, but then we have an intriguing side effect: Since every neighborhood around any other point also contains $p$, said sequence also converges against every other point as well!

Now we have some interesting questions:

  1. Can we “add” a point which is a companion point to only a subset of our original space?
  2. If we add two companion points, are they automatically topologically indistinguishable?
  3. Can we construct $\comp{\mathbb R}$ as a pushout or some other “universal” construction?

Companionship: The specialization preorder

The point we added to our topology appears to be a kind of extreme example of what may appear in a weaker fashion: With respect to any different point, every open set containing the latter contains our new point as well – which suggests the following definition:

DEF Let $X$ be a topological space. $c\in X$ is said to “accompany” (or “be a companion to”) $x\in X$ if every open set containing $x$ also contains $c$. Analogously, $x$ is said to “accompany” $S\subseteq X$ if it accompanies every point in $S$. In symbols, we will use $c \sphericalangle x$ or $c \sphericalangle S$ (evaluated pointwise).

The following consequences are immediate:

  1. Companionship is transitive and reflexive, thus amounts to a preorder.
  2. If $x\sphericalangle y$ and $y\sphericalangle x$, those points are topologically indistinguishable (and vice versa). In particular, if our space satisfies the $T_0$ separation axiom, that is, every distinct points are topologically distinguishable, $x\sphericalangle y$ and $y\sphericalangle x$ implies $x=y$, in other words, we have a poset.
  3. If our space is $T_1$, meaning that given two points $x,y$, each of them can be separated by an open set from the other, we directly conclude that neither $x\sphericalangle y$ nor $y\sphericalangle x$ can hold, so any two distinct points are incomparable – meaning our companionship relation is precisely the diagonal.

The latter motivates our order as a nice tool to investigate how “far” a space is from being $T_1$.

For later purposes, we show that it is sufficient to check companionship on a basis (the proof is straightforward):

LEM Let $c, s\in X$, $\mathscr B$ a basis of $X$. If every $B\in \mathscr B$ which contains $s$ also contains $c$, $c$ is a companion to $s$. Furthermore, if $S\subseteq X$ and every $B\in \mathscr B$ which is not disjoint to $S$ also contains $c$, $c$ accompanies $S$.

Proof Let $s\in U = \bigcup_{i\in I}B_i$. This means that for some $i$, $s\in B_i$, and by our condition, we also have $c\in B_i\subseteq U$. For the second part, fix an $s\in S$. We need to show that $c$ is a companion to $s$: Let $s\in U = \bigcup_{i\in I} B_i$, so again, $s\in B_i$ for some $i\in I$. Containing $s$ means in particular not being disjoint to $S$, so by our premise, $c\in B_i\subset U$, and we're finished.

Here comes the shameful part where I realize that the order I defined not only already exists, but was already known to me. It is called the “specialization preorder” – however, usually it is defined by the following, equivalent characterization:

LEM $\DeclareMathOperator{cl}{cl}$Let $x, y\in X$. $\cl(x)\subseteq \cl(y) \Leftrightarrow y\sphericalangle x$.

Proof First, we will reformulate the right hand side to deal with closed, and not open sets: \begin{align} y\sphericalangle x \Leftrightarrow&\forall U \:\text{open}\colon\: x\in U\implies y\in U\\ \Leftrightarrow&\forall U \:\text{open}\colon\: y\not\in U\implies x\not\in U\\ \Leftrightarrow&\forall U \:\text{open}\colon\: y\in U^c\implies x\in U^c\\ \Leftrightarrow&\forall C \:\text{closed}\colon\: y\in C\implies x\in C\\ \end{align} Now, we will remember that $\cl$ is a closure operator and satisfies extensitivity, monotony, and idempotence, so if $x\in C$, writing this as $\{x\}\subseteq C$ and applying $\cl$ to both sides, we get $\cl(x)\subseteq \cl(C)=C$, and on the other hand, because $x\in \cl(x)$, the latter implies $x\in C$ as well, so we have an equivalence: \begin{align} &\forall C \text{ closed}\colon\: y\in C\implies \cl(x)\subseteq C\\ \Leftrightarrow&\forall C \text{ closed}, y\in C \colon\: \cl(x)\subseteq C\\ \Leftrightarrow& \cl(x)\subseteq \bigcap_{\substack{y\in C \\ C\:\text{closed}}} C = \cl(y)\\ \end{align} Which completes the proof.

In most scenarios, I find it more intuitive to describe topologies with respect to their open sets, but that is just the personal preference of a layman.

Note that we can make the following statements about points:

  1. Let $\{x\}$ be open. Then $y\sphericalangle x$ implies in particular that $y\in \{x\}$, i.e. $y=x$ – in other words, open points are minimal.
  2. Let $\{x\}$ be closed. With the dual characterization, $x\sphericalangle y \Leftrightarrow \cl(y)\subseteq \cl(x) = \{x\}$, and since $y\in \cl(y)$, we also have that $y=x$: closed points are maximal!

A useful thing to keep in mind is that when two are maximal (respectively, minimal), they are either equal or incomparable.

Other ideas (WIP)

DEF Let $X$ be a topological space. Define $\operatorname{Comp}(X, S, p)$ to be the set $X\sqcup {p}$ equipped with the topology generated by the following set: \begin{align} \mathscr T &:= \langle \mathscr A \cup \mathscr A^\prime \cup \{\{p\}\} \rangle, \end{align} where \begin{align} \mathscr A &:= \{ O\cup \{p\} \mid O\cap S \neq \emptyset, O \ \text{open in }\ X \} \\ \mathscr A^\prime &:= \{ O \mid O\cap S = \emptyset, O \ \text{open in }\ X \} \end{align}

DEF Let $(X,\mathscr T_X), (Y, \mathscr T_Y)$ be a topological spaces, $S\subseteq Y$. Define a topological space $$ \comp(X,S,Y) := \{ X\sqcup Y, \mathscr T_X \cup \Sigma \cup \mathscr R \} $$ where \begin{align} &\Sigma := \{ X \cup O \mid O\in \mathscr T_Y,\,O\cap S\neq \emptyset \},\\ &\mathscr R := \{ O \mid O\in \mathscr T_Y,\,O\cap S=\emptyset \} \end{align}

LEM The generating set given above is a basis.

Proof We wil verify that pairwise intersections can be represented as unions of our generator. To do that, consider the following cases separately:

LEM I

Proof We will verify that the generating set given in the above definition forms a basis.

Q What do companions to the whole space mean for the preorder / depth associated to a finite topology?

Q Is a point that's the only minimum in its connected component necessarily open?

Comparing more things than points

As mentioned, as soon as we have $T_1$, the specialization preorder is discrete, thus does not convey any more information. But what if we compare subsets of a partition in a similar way? Take $\mathbb R = (-\infty, 0) \cup [0,\infty)$. We can “represent” this partition by the homomorphic image onto a sierpinski space. There, the specialization preorder reflects that it is not discrete, i.e. that partition is not a partition into closed sets. Questions: