Nowhere Dense Sets in Discrete Metric Space

Question

Not Homework- Just Personal Study

Let $(M,d)$ be a discrete metric space. I want to show that the only subset of $M$ that is nowhere dense is the empty set. Let $E$ be a non-empty subset of $M.$ The definition of nowhere dense I have access to is that $E \subset M$ is nowhere dense if it is not dense in any open ball contained in $M.$ This is the same as saying every open ball must contain an interior point. I am stuck here. How do I know that there is an open ball in my space that contains a point not in $E$? I am sure I am over thinking I am just missing this final piece, or perhaps I am missing something obvious. I do not have access to the definition of nowhere dense which takes a set to be nowhere dense if the interior of its closure is empty as the book I am using (Kolmogorov and Fomin) does not provide this definition.

Edit 1: I think I have answered my question, but I want to clarify that my answer is correct. I first wanted to introduce the definitions I am working with to make things precise. Kolmogorov and Fomin - Introductory Analysis.

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Definition 1: Let $(X,d)$ denote the parent metric space. Let $A,B$ be subsets of $X.$ $A$ is said to be dense in $B$ if $B \subset \overline{A}.$

Definition 2: Let $(X,d)$ denote the parent metric space. Then $A\subset X$ is everywhere dense in $X$ if $\overline{A}=X.$

Definition 3: Let $(X,d)$ denote the parent metric space. Then $A$ is said to be nowhere dense if it is dense in no open ball at all.

We use these definitions to prove the following result.

Proposition: Let $(X,d)$ be a discrete metric space. Then the only nowhere dense set in $(X,d)$ is the empty set.

Proof: We first show that it is in fact the case that the empty set is nowhere dense in this space. Let $B(x,\epsilon)$ be an open ball in $(X,d).$ Then the result is clear, because the empty set is its own closure and cannot contain a non empty set. Now take any non empty subset $E$ in $(X,d).$ To show that this set is not nowhere dense, we must show that $E$ is dense in some open ball. Take a point $y \in E.$ Then $\overline{E}=E$ and

$$\{y\}=B(\frac{1}{2},y) \subset E=\overline{E}$$

Thus, $E$ is dense in $B(\frac{1}{2},y),$ and is thus not nowhere dense in $(X,d).////$

My original confusion was not understanding the distinction between dense and everywhere dense, and now, that containing an open ball is enough is clear. Does the argument look OK?

Thank you again.

Answer 1

If $M$ is discrete, every subset of $M$ is open. If $E$ is a nonempty subset of $M$ and $p \in E$, then the singleton $\{p\}$ is open, which says there exists $\delta > 0$ such that the ball of radius $\delta$ around $p$ contains no points other than $p$. In particular, $E$ is dense in that ball.

Answer 2

A very technical proof using the following characterisation of nowhere dense sets: $A$ is said to be nowhere dense in $X$ if the closure of $A$ has empty interior.

Now if $X$ is discrete then every subset is closed and open so $A$ closure is $A$ whereas $A$ interior is $A$ minus the closure of its complement i.e. closure of $X - A$ but then $X - A$ is closed as well so $A$ closure interior is just $A$ which is empty only when $A$ is empty.