- #1

Eclair_de_XII

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- TL;DR Summary
- Any bounded, infinite subset of the real numbers must contain at least one limit point.

Let ##X## be a bounded subset of ##\mathbb{R}## with infinite cardinality. We consider a countably-infinite subset of ##X##. We write this set as a sequence to be denoted ##\{a_n\}_{n\in\mathbb{N}}##.

Now define ##A## to be the set of points in the sequence with the property that for each ##a_n\in A##, there is an open set that owns ##a_n## with the property that the intersection of the open set w.r.t. to the sequence is empty. If this condition is fulfilled, then choose the open set ##U_n## s.t. there is a point ##a_k\in\{a_n\}_{n\in\mathbb{N}}## s.t. ##a_k=\sup U_n##.

Suppose ##A^C=\emptyset## and assume ##\inf\{a_n\}_{n\in\mathbb{N}}\in A##. Suppose ##\sup\{a_n\}_{n\in\mathbb{N}}\in A##. By definition, there is a point ##a_k## s.t. ##a_k>\sup\{a_n\}_{n\in\mathbb{N}}##, which is a contradiction. Hence, ##\sup\{a_n\}_{n\in\mathbb{N}}\in A^C##, contrary to supposition.

Now define ##A## to be the set of points in the sequence with the property that for each ##a_n\in A##, there is an open set that owns ##a_n## with the property that the intersection of the open set w.r.t. to the sequence is empty. If this condition is fulfilled, then choose the open set ##U_n## s.t. there is a point ##a_k\in\{a_n\}_{n\in\mathbb{N}}## s.t. ##a_k=\sup U_n##.

Suppose ##A^C=\emptyset## and assume ##\inf\{a_n\}_{n\in\mathbb{N}}\in A##. Suppose ##\sup\{a_n\}_{n\in\mathbb{N}}\in A##. By definition, there is a point ##a_k## s.t. ##a_k>\sup\{a_n\}_{n\in\mathbb{N}}##, which is a contradiction. Hence, ##\sup\{a_n\}_{n\in\mathbb{N}}\in A^C##, contrary to supposition.

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