**Finite Dense Subsets**

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

Let be a metric space. Suppose is a finite dense subset. Then, that must also be finite.

As a direct proof, this is proven quite easily by showing that since , so that .

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 be a set with supremum . Adjoin an element to , . so that . Then, .
- Use this, and proceed by induction to see that all finite sets 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 is infinite. We will show that must also be infinite. Let . Since , for , there is , . Now, take . Either or not. In any case, set . Then, . But, since is dense, we must have . Inductively, take so that either or In either case, set which exists for every . Moreover, for . But, since is again dense, we must have . Repeating this process, we see that must be at if is at least . So, we must have finite if is a finite dense subset.

Cheers,

J.T. (S.D.)

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