# A Note on Finite Dense Subsets

Finite Dense Subsets

This will be a short entry, just a little note on the cardinality of dense subsets. That is,

Let $(X,d)$ be a metric space. Suppose $E \subset X$ is a finite dense subset. Then, that $X$ must also be finite.

As a direct proof, this is proven quite easily by showing that $\bar{E} = E$ since $E' = \emptyset$, so that $\bar{E} = X$.

More generally, this stems from the fact that all finite sets contain their supremums (and therefore their limit points). The best way to prove this is with the following argument:

• Let $X \subset Y$ be a set with supremum $\beta$.  Adjoin an element to $X$, $\alpha \in Y$. so that $X' = X \cup \{\alpha\}$. Then, $\sup\{X'\} = \max\{\beta,\alpha\}$.
• Use this, and proceed by induction to see that all finite sets $X$ contain their supremums, and thus have empty limit point sets.

However, I find this proof a bit boring, as sometimes direct proofs can be. So, to spice things up a little bit (because hey math can be sexy too), here’s an original proof by contradiction:
Suppose $X$ is infinite. We will show that $E$ must also be infinite. Let $x_1 \in X$. Since $E \overset{dense}{\subset} X$, for $\epsilon >0$, there is $p_1 \in B_\epsilon (x_1)$, $p_1 \in E$. Now, take $x_2 \in X$. Either $x_2 \in B_\epsilon (x_1)$ or not. In any case, set $r_2 = \frac{1}{2} d(x_2,p_1)$. Then, $p_1 \notin B_{r_2} (x_2)$. But, since $E$ is dense, we must have $p_2 \in B_{r_2} (x_2)$. Inductively, take $x_k \in X$ so that either $x_k \in B_\epsilon(x_1)$ or $x_k \in \bigcup \limits_{i=1}^{k-1} B_{r_i} (x_i).$ In either case, set $r_k = \min\{d(x_k,p_1), \ldots, d(x_k,p_{k-1})\}$which exists for every $k$. Moreover, $p_j \notin B_{r_k}(x_k)$ for $j\leq k-1$. But, since $E$ is again dense, we must have $p_k \in B_{r_k}(x_k)$. Repeating this process, we see that $|E|$ must be at $\aleph_0$ if $|X|$ is at least $\aleph_0$. So, we must have $X$ finite if $E$ is a finite dense subset.

Cheers,

J.T. (S.D.)