Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\operatorname{cl}$ is the closure operator for some topology.

I will call induced proximity the proximity defined by the formula:

$$A\delta B\Leftrightarrow \operatorname{cl}(A)\cap\operatorname{cl}(B)\ne\varnothing.$$

Is induced proximity really a proximity for every given topological space?

Also: What I call here induced proximity is the weakest proximity generating our topology, right?

share|cite|improve this question
up vote 1 down vote accepted

From Wikipedia, I learn about proximities:

The resulting topology is always completely regular.

Thus either induced proximity fails to be a proximity if the given topological space is not regular. Or the generated topology may differ from the given topology.

The latter kind of failure occurs in the space $\{1,2\}$ with open sets $\emptyset$, $\{1\}$, $\{1,2\}$. Here, $\{2\}$ is closed in the given topology, but in the generated topology, the closure of $\{2\}$ is $\{x\mid \{x\}\delta\{2\}\}=\{1,2\}$ (because $\operatorname{cl}(\{1\})=\{1,2\}$).

share|cite|improve this answer

$\newcommand{\cl}{\operatorname{cl}}$Check the axioms from this answer. It’s immediate that $P_0$ and $P_2$ are satisfied. Suppose that $A\delta(B\cup C)$; then $\varnothing\ne\cl A\cap\cl(B\cup C)=\cl A\cap\big(\cl B\cup\cl C\big)=(\cl A\cap\cl B)\cup(\cl A\cap\cl C)$, so at least one of $\cl A\cap\cl B$ and $\cl A\cap\cl C$ is non-empty, and $A\delta B$ or $A\delta C$; thus, $P_1$ is satisfied.

$P_3$, however, becomes $\cl\{x\}\cap\cl\{y\}=\varnothing$ iff $x\ne y$ in this setting; for this you want $X$ to be $T_1$.

$P_4$ is also problematic. If $A\bar\delta B$, then $\cl A\cap\cl B=\varnothing$. We want to find $C,D\subseteq X$ such that $A\bar\delta C$, $B\bar\delta D$, and $C\cup D=X$, i.e., such that $\cl A\cap\cl C=\cl B\cap\cl D=\varnothing$ and $C\cup D=X$. Setting $U=X\setminus\cl C$ and $V=X\setminus\cl D$, we see that this requires finding open sets $U,V\subseteq X$ such that $\cl A\subseteq U$, $\cl B\subseteq V$, and $U\cap V=\varnothing$, so for this we want $X$ to be normal.

You do get a proximity if $X$ is $T_4$.

share|cite|improve this answer
According to Wikipedia, $P_3$ is $A \cap B \neq \emptyset \implies A\delta B$. Since $A \cap B \subseteq \text{cl}(A \cap B) \subseteq \text{cl}(A) \cap \text{cl}(B)$ it follows that if $A \cap B \neq \emptyset$ then $\text{cl}(A) \cap \text{cl}(B) \neq \emptyset$ and so $A\delta B$. Why is $T_1$ necessary? – Fly by Night Oct 22 '12 at 16:40
@FlybyNight: There is a reason that I linked to this answer: I’m using the same axioms. With a slight renumbering they are the axioms given in Engelking’s General Topology and with yet another numbering, the axioms to be found in Willard’s General Topology. – Brian M. Scott Oct 22 '12 at 16:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.