Question

1.5.7 Prove the following separately Theorem 1.5.10. Let (X,d) be a metric space. (a) IfY is a compact subset of X, and Z C Y, then Z is compact if and only if Z is closed (b) IfY. Y are a finite collection of compact subsets of X, then their union Y1 U...UYn is also compact. (c) Every finite subset of X (including the empty set) is compact.

Answer 1

Proof (a) Before we prove this, certain concepts and results need to be established. For a compact metric space, the properties of Bolzano - Werestras and sequentially compact are equivalent to compactness sequentially compact is essentially every sequence in the space has a convergent subsequence Coming to closed sets, a set is closed if there exists a sequence {x_n} converging to the limit point in the set it self If z subsetorequalto y is compact, then it is also sequentially compact. I . e every sequence in z has a convergent subsequence. Now if x elementof y was a limit point of z. \r\n then there asserts a sequence {x_n} in z, s . t x_n rightarrow x But we also have a convergent subsequence {x_n k}, by being sequentially compact. Which means that x_nk rightarrow k and x elementof z Hence x is closed suppose z is closed. Let {H_z} be an open cover of z. the corresponding open sets {G_1}