Bolzano-Weierstrass Theorem: *Every bounded, infinite set of real numbers has a limit point.*

**Lemma**

Let S be a set with endowed with a linear order such that for all nonempty , both and exists and belong to S. Then S is finite.

**Proof of the lemma:** Suppose that S satisfies the linear order and it is infinite. For , we can recursively construct and . Then is an infinitely increasing sequence of member of S. Hence, which is a contradiction.

**Proof of Theorem:**

By contraposition. Let S be a bounded subset of and assume it has no limit point. Suppose , as constructed above, is nonempty. Then . Note that is a limit point of X, hence of S too. Analogously, . Then by Lemma , S is finite. [1]