note-math

compact's original inspiration: The intersection of closed interval nets of $\mathrm{ℝ}$ is non-empty #link(<closed-interval-net-theorem>)[]

$0$ is a #link(<limit-point>)[limit point] of ${\text{T}}_{\mathrm{ℝ}}$ of $\left(0,1\right)$. The #link(<net>)[net] $\left(0,\frac{1}{𝑛}\right)$ seems to converge to $0$

But $\bigcap \left(0,\frac{1}{𝑛}\right)=\varnothing$

Compare the multiplicative inverse's $\bigcap \left(𝑛,\infty \right)=\varnothing$

$\mathrm{ℝ}$ does not have $\infty$ corresponding to the possible limit $0$

Compare $\bigcap \left[0,\frac{1}{𝑛}\right)=\left\{0\right\}$

let ${\text{T}}_{𝑋}$ #link(<topology>)[topological space]. let $𝐴\subset 𝑋$

compact_(tag) $𝐴$ compact := forall $\text{B}$ net of $𝐴$, ${\bigcap }_{𝐵\in \text{ B}}\overline{𝐵}\ne \varnothing$

Meaning: The elements of any #link(<net>)[net] $\text{B}$ have a common limit point set under the topology ${\text{T}}_{𝑋}$. Or, after ${\text{T}}_{𝑋}$ closure, the net $\text{B}$ converges to a non-empty set or the intersection is non-empty, instead of converging to the empty set (e.g., Euclidean ${\mathrm{ℝ}}^{𝑑}$ converges to the empty set or converges to infinity, but there are many other complex situations)

Any net can replenish all finite intersections and maintain #link(<net-same-limit>)[the same limit], so for compact, the equivalent description is

${\text{T}}_{𝑋}$ compact <==>

$$\forall {𝐴}_1,\dots ,{𝐴}_{𝑛}\in \text{ A},{\overline{𝐴}}_1\cap \cdots \cap {\overline{𝐴}}_{𝑛}\ne \varnothing ⟹\bigcap _{𝐴\in \text{ A}}\overline{𝐴}\ne \varnothing$$

logically equivalent to

$$\bigcap _{𝐴\in \text{ A}}\overline{𝐴}=\varnothing ⟹\exists {𝐴}_1,\dots ,{𝐴}_{𝑛}\in \text{ A},{\overline{𝐴}}_1\cap \cdots \cap {\overline{𝐴}}_{𝑛}=\varnothing$$

logically equivalent to compact-finite-open-cover_(tag)

$$\bigcup _{𝐴\in \text{ A}}\mathring{𝐴}=𝑋⟹\exists {𝐴}_1,\dots ,{𝐴}_{𝑛}\in \text{ A},{\mathring{𝐴}}_1\cap \cdots \cap {\mathring{𝐴}}_{𝑛}=𝑋$$

compact-subset_(tag) $𝑆\subset 𝑋$ := #link(<topology-subspace>)[] ${\text{T}}_{𝑆}$ compact

recall #link(<closed-in-subspace>)[], $\text{closed}\left(𝐴,{\text{T}}_{𝑆}\right)=𝑆\cap \text{closed}\left(𝐴,{\text{T}}_{𝑋}\right)$, denoted as $𝑆\cap \overline{𝐴}$

compact-subset logically equivalent to

$$\forall {𝐴}_1,\dots ,{𝐴}_{𝑛}\in \text{ A},𝑆\cap {\overline{𝐴}}_1\cap \cdots \cap {\overline{𝐴}}_{𝑛}\ne \varnothing ⟹𝑆\cap \bigcap _{𝐴\in \text{ A}}\overline{𝐴}\ne \varnothing$$

logically equivalent to

$$𝑆\cap \bigcap _{𝐴\in \text{ A}}\overline{𝐴}=\varnothing ⟹\exists {𝐴}_1,\dots ,{𝐴}_{𝑛}\in \text{ A},𝑆\cap {\overline{𝐴}}_1\cap \cdots \cap {\overline{𝐴}}_{𝑛}=\varnothing$$

logically equivalent to compact-subset-finite-open-cover_(tag)

$$𝑆\subset \bigcup _{𝐴\in \text{ A}}\mathring{𝐴}⟹\exists {𝐴}_1,\dots ,{𝐴}_{𝑛}\in \text{ A},𝑆\subset {\mathring{𝐴}}_1\cap \cdots \cap {\mathring{𝐴}}_{𝑛}$$

compact-subset is closed under finite unions. this is easy to proof

closed-set-in-compact-space-is-compact_(tag) ${\text{T}}_{𝑋}$ compact and $𝑆$ closed ==> $𝑆$ compact

Proof

Hausdorff space := $\forall 𝑥,{𝑥}^{\prime }\in 𝑋,𝑥\ne {𝑥}^{\prime }⟹\exists {𝐵}_{𝑥},{𝐵}_{{𝑥}^{\prime }}\in {\text{ T}}_{𝑋},{𝐵}_{𝑥}\cap {𝐵}_{{𝑥}^{\prime }}=\varnothing$

Hausdorff + compact ==> closed. At this time, compact is closed for any intersection

continous-preserve-compact_(tag) let $𝑓:𝑋\to 𝑌$. $𝑓\left(𝑋\right)$ is compact-subset of ${\text{T}}_{𝑌}$

Proof

Contrapositive: Under a continuous function, the inverse image of non-compact is non-compact

quotient-topology-preserve-compact_(tag) For #link(<quotient-topology>)[] $𝜋:𝑋⇝\frac{𝑋}{\sim }$, source space $𝑋$ compact ==> quotient space $\frac{𝑋}{\sim }$ compact. because the quotient map $𝜋$ is continuous, it preserves compact

product-topology-preserve-compact_(tag) #link(<product-topology>)[] preserves compact

Proof