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.


J.T. (S.D.)


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