- #1

Eclair_de_XII

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- 91

- TL;DR Summary
- Any closed and bounded subset of the set of real numbers must be compact.

*Need someone to check my work and reasoning.

Let ##X## be a closed and bounded subset of the real numbers. Let ##\{x_i\}_{i\in I}##, for some index ##I##, represent the set of limit points of ##X##. Since ##X## is closed, it must follow that ##\{x_i\}_{i\in I}\subset X##. Hence, the set of limit points must be bounded.

Let ##\{U_j\}_{j\in J}## denote an open cover for ##X'##, where ##J## is some index. This open cover must own an open set ##U_0## s.t. ##\inf X'\in U_0##. Suppose there is ##x_i\in X'## s.t. ##x_i\notin U_0##. Then there must exist ##U_i\in\{U_j\}_{j\in J}## s.t. ##x_i\in U_i##. Let ##u_i:=\sup(U_i)##. Continuing in this fashion, we obtain a collection of open sets ##\{U_k\}_{k\in K}## that cover ##X'##, where ##K## is a subset of the natural numbers.

Assume ##\{U_k\}_{k\in K}## owns the minimal amount of elements and suppose this open cover has infinite cardinality. Then ##\{u_k\}_{k\in K}## is an increasing sequence of points that converges to its supremum, to be denoted ##u_0##.

Necessarily, there is an open set ##U'\in\{U_k\}_{k\in K}## s.t. ##u_0\in U'##. Let ##\epsilon>0## be small enough s.t. ##B(u_0,\epsilon)\subset U'##. Then there is a natural number ##N## s.t. whenever ##n\in\mathbb{N}## and ##n\geq N##, ##u_n\in B(u_0,\epsilon)\subset U'##. The open set ##U'## contains the set ##\{u_n:n\geq N\}##, and each of the ##N-1## sets of the form ##U_i## own ##u_i##. Hence, these ##N-1## covers ##U_i## and ##U'## are sufficient in order to cover ##X'##, contrary to the assumption that ##\{U_k\}_{k\in K}## is the smallest cover with this property.

Let ##\{U_j\}_{j\in J}## denote an open cover for ##X'##, where ##J## is some index. This open cover must own an open set ##U_0## s.t. ##\inf X'\in U_0##. Suppose there is ##x_i\in X'## s.t. ##x_i\notin U_0##. Then there must exist ##U_i\in\{U_j\}_{j\in J}## s.t. ##x_i\in U_i##. Let ##u_i:=\sup(U_i)##. Continuing in this fashion, we obtain a collection of open sets ##\{U_k\}_{k\in K}## that cover ##X'##, where ##K## is a subset of the natural numbers.

Assume ##\{U_k\}_{k\in K}## owns the minimal amount of elements and suppose this open cover has infinite cardinality. Then ##\{u_k\}_{k\in K}## is an increasing sequence of points that converges to its supremum, to be denoted ##u_0##.

Necessarily, there is an open set ##U'\in\{U_k\}_{k\in K}## s.t. ##u_0\in U'##. Let ##\epsilon>0## be small enough s.t. ##B(u_0,\epsilon)\subset U'##. Then there is a natural number ##N## s.t. whenever ##n\in\mathbb{N}## and ##n\geq N##, ##u_n\in B(u_0,\epsilon)\subset U'##. The open set ##U'## contains the set ##\{u_n:n\geq N\}##, and each of the ##N-1## sets of the form ##U_i## own ##u_i##. Hence, these ##N-1## covers ##U_i## and ##U'## are sufficient in order to cover ##X'##, contrary to the assumption that ##\{U_k\}_{k\in K}## is the smallest cover with this property.