# general topology – Set inclusion for reals

Statement: $$K subset mathbb R$$ is compact implies $$K$$ is closed.

Proof: Suppose $$K$$ is compact. Consider the collection of open sets $$displaystyle{O_{k_{in mathbb N}} = left{x in mathbb R: |x – a| > frac 1kright}}$$. Then $$color{green}{displaystyle{bigcup_{k=1}^infty O_k = mathbb R setminus {a}}}$$ and so $$displaystyle{K subsetbigcup_{k=1}^infty O_k}$$. Since $$K$$ is compact, $$displaystyle{K subsetbigcup_{j=1}^m O_{k_j}}$$. Let $$N = max{k_i}_{i=1}^m$$. Then $$displaystyle{K subsetbigcup_{j=1}^m O_{k_j} = left{x in mathbb R: |x – a| > frac 1Nright}}$$. By taking complements, we have $$displaystyle{left{x in mathbb R: |x – a| < frac 1Nright} subset left{x in mathbb R: |x – a| color{red}{le} frac 1Nright} color{blue}{subset} K^c}$$ meaning $$a$$ is an interior point of $$K^c$$ implying $$K^c$$ is open which to say $$K$$ is closed.

I’d like to prove the relations in $$color{green}{text{green}}$$ and $$color{blue}{text{blue}}$$ and give a reason for the one in $$color{red}{text{red}}$$ above. Please, see if what I typed below works. Thanks.

Statement: $$color{green}{displaystyle{bigcup_{k=1}^infty O_k = mathbb R setminus{a}}}$$.

Proof:

Note, $$a not in O_k$$ because $$|a – a| not > frac 1k$$ for any $$k$$.

Suppose $$displaystyle{x in bigcup_{k=1}^infty O_k}$$. Then $$x in O_k$$ for some $$k$$ and so $$x in mathbb R setminus {a}$$ by definition of $$O_k$$. Now assume $$x in mathbb R setminus {a}$$. By Archimedes, there’s some $$K in mathbb N$$ s.t. $$displaystyle{K > frac{1}{|x – a|} iff frac 1K < |x – a|}$$ and so $$x in O_K$$ meaning $$displaystyle{x in bigcup_{k=1}^infty O_k}$$.

Statement: $$color{blue}{displaystyle{left{x in mathbb R: |x – a| le frac 1N right} subset K^c.}}$$

Proof:

Since $$K subset mathbb R$$, we have that $$K^c = mathbb R setminus K.$$ Suppose $$displaystyle{s in left{x in mathbb R: |x – a| le frac 1N right}}$$. Then $$displaystyle{|s – a| le frac 1N}$$. Since $$displaystyle{K subset bigcup_{j=1}^mO_{k_j}}$$, every element $$t in K$$ has the property $$displaystyle{|t – a| > frac 1k}$$ and so $$s ne t$$ meaning $$s not in K$$ implying $$s in K^c$$.

We need $$displaystyle{left{x in mathbb R: |x – a| le frac 1N right} subset K^c}$$ instead of $$displaystyle{left{x in mathbb R: |x – a| < frac 1N right} subset K^c}$$ because $$O_k^c = displaystyle{left{x in mathbb R: |x – a| le frac 1k right}}$$ by trichotomy law for reals.