# Metrics and the Kuratowski closure axioms

Edit: Succinct proofs from user87690 can be found below, but I will gladly up-vote other valid approaches to any of the problems here!

The following questions concern closure operators and the Kuratowski closure axioms.

Additional tags and clarifications for the problems are all welcome.

Fact: Let $X$ be a nonempty set with $Y ⊂ X$. Define $\textbf{K}$ as follows: for any $A ⊂ X$, let $\textbf{K}A = A$ if $A$ is finite, and $A \cup Y$ otherwise. Then $\textbf{K}$ is a closure operator on $X$.

(Motivation: These are original problems; I am wondering (a) if they have already appeared elsewhere, and (b) how one could resolve them with an emphasis on parsimony.)

i) Let $X = Y = \mathbb{Z}^+$. Define $\textbf{K}$ as above. Does this closure operator correspond to a metric? If so, give such a metric; if not, provide a proof.

ii) Let $X = \mathbb{Z}^+$ and let $Y = \{5\}$. Define $\textbf{K}$ as above. Does this closure operator correspond to a metric? If so, give such a metric; if not, provide a proof.

iii) Let $X = \mathbb{R}$ and let $Y = \{5\}$. Define $\textbf{K}$ as above. Does this closure operator correspond to a metric? If so, give such a metric; if not, provide a proof.

Ideally, the proofs would be in the language of the closure axioms, rather than translating into standard topological notation (in the sense of, e.g., Munkres). If memory serves, then the only closure operator that corresponds to a metric is the second one listed above.

In i) all finite sets are closed and all infinite sets are dense. This is a characterization of cofinite topology on a set, which is minimal $T_1$ topology. On infinite set it is not even $T_2$ and hence non-metrizable.

In ii) and iii) we get one-point compactification of discrete space on $X \setminus Y$. Such space is metrizable iff it is countable, which is the case of ii). There the space is just a convergent sequence.

Since I am not familiar with arguments using one-point compactification, here is what I tried to use for questions two and three.

(ii) Consider $d: \mathbb{Z}^+ \times \mathbb{Z}^+ \rightarrow \mathbb{R}$ as defined below.

Given $m, n \in \mathbb{Z}^+$, we compute $d(m,n)$ as follows:

$|1/(m-5)|$ iff only $n = 5$;

$|1/(n-5)|$ iff only $m = 5$;

$0$ iff $m = n = 5$;

$|1/(m-5) - 1/(n-5)|$ otherwise.

Then $d$ satisfies the axioms for a metric, and induces the desired topology. Q.E.D.

(iii) Suppose for the sake of contradiction that $\textbf{K}$ is induced by a metric $d$.

Define $C_n := \{x \in \mathbb{R}: d(x,5) \geq 1/n\}$.

Observe each $C_n$ is finite; so $\mathbb{R} - \{5\} = \bigcup\limits_{n \in \mathbb{Z}^{+}} C_n$ is countable; contradiction.

Thus, we must have been wrong in our supposition, and $(\textbf{K}, \mathbb{R})$ is not a metric space. Q.E.D.

• Note that your proof of (iii) works when proving that one-point compactification of uncountable discrete space is not first-countable at the added point, and hence is not metrizable. Using the wording “one-point compactification of uncountable discrete space” can be seen just as a fast and robust way of specifying the topology we are talking about. – user87690 Oct 26 '14 at 11:10
• Also note that in (ii), the particular structure of our space is rather an obstacle to prove the desired properties. It may be better to forget it. Let's take the space $\{0\} ∪ \{\frac{1}{n}: n ∈ \mathbb{N}\} ⊆ \mathbb{R}$. It is obviously metrizable and its topological characterization corresponds to the characterization of our space (“the convergent sequence space”). Note the ordering of the isolated points doesn't matter, we may map the limit point to the limit point and take any bijection between the isolated points and we get a homeorphism, which may transfer the metric. – user87690 Oct 26 '14 at 11:28
• @user87690 Thanks for the comments and answer; it is quite nice to see a new way of looking at a problem, and not only the new ways of thinking that come with it, but also the new vocabulary. – Benjamin Dickman Oct 26 '14 at 21:56