Finite Union of Compact Sets is Compact

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $n \in \N$ be a natural number.

Let $\sequence {U_i}_{1 \mathop \le i \mathop \le n}$ be a finite sequence of compact subsets of $T$.

Let $\UU_n := \ds \bigcup_{i \mathop = 1}^n U_i$ be the union of $\sequence {U_i}$.


Then $\UU_n$ is compact in $T$.


Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\UU_n := \ds \bigcup_{i \mathop = 1}^n U_i$ is compact in $T$.


\map $P 0$ is the case:

$\UU_0 := \ds \bigcup_{i \mathop = 1}^0 U_i$

From Union of Empty Set:

$\ds \bigcup_{i \mathop = 1}^0 U_i = \O$

From Empty Set is Compact Space it follows that:

$\UU_0$ is compact in $T$.


$\map P 1$ is true, as this just says:

$U_1$ is compact in $T$.


Basis for the Induction

$\map P 2$ is the case:

$U_1 \cup U_2$ is compact in $T$.

This is proved in Union of Two Compact Sets is Compact.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\UU_k := \ds \bigcup_{i \mathop = 1}^k U_i$ is compact in $T$.


Then we need to show:

$\UU_{k+1} := \ds \bigcup_{i \mathop = 1}^{k + 1} U_i$ is compact in $T$.


Induction Step

This is our induction step:

We have that:

$\ds \bigcup_{i \mathop = 1}^{k + 1} U_i = \paren {\bigcup_{i \mathop = 1}^k U_i} \cup U_{k + 1}$

By the induction hypothesis:

$\UU_k$ is compact in $T$.

By the basis for the induction:

$\UU_k \cup U_{k + 1}$ is compact in $T$.


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\ds \forall n \in \N: \bigcup_{i \mathop = 1}^n U_i$ is compact in $T$.

$\blacksquare$