note-math

Example #link(<interval>)[] #link(<best-interval-decomposition>)[]

connected_(tag) Connected or limit point connected := limit point set decomposition is no longer possible or closed set decomposition is no longer possible

$𝑋={⨆}_{𝑖\in 𝐼}{𝐴}_{𝑖}$ with ${𝐴}_{𝑖}$ closed ==> $\left|𝐼\right|=1$

Intuitively speaking, connected = cannot give any substantial decomposition. For any set decomposition $𝑋=⨆{𝐴}_{𝑖}$, from $𝑋=\bigcup {\overline{𝐴}}_{𝑖}$ + connected, each ${𝐴}_{𝑖}$ must be connected to some other ${𝐴}_{𝑗}$ after the limit point: $\exists 𝑗\ne 𝑖,{\overline{𝐴}}_{𝑖}\cap {\overline{𝐴}}_{𝑗}\ne \varnothing$

Each ${𝐴}_{𝑖}$ of the closed set decomposition is an open set

Proof ${𝐴}_{𝑖}=𝑋\setminus \overline{{\bigcup }_{{𝑖}^{\prime }\ne 𝑖}{𝐴}_{{𝑖}^{\prime }}}$

The definition of connected is equivalent to the version decomposed into two closed sets

$𝑋=𝐴\bigsqcup {𝐴}^{\prime }$ with $𝐴,{𝐴}^{\prime }$ closed ==> $\left(𝑋=𝐴\right)\vee \left(𝑋={𝐴}^{\prime }\right)$

Proof Taking the limit of the decomposition yields ${\bigvee }_{𝑖\in 𝐼}𝑋={𝐴}_{𝑖}$

Connected subset := topological subspace connected

Example $\mathrm{ℝ}$ is connected. $\mathrm{ℝ}$ has connected and disconnected sets. Connected sets may not be ${\text{T}}_{\mathrm{ℝ}}$ closed sets

real-connected-is-interval_(tag) Connected sets of $\mathrm{ℝ}$ are intervals Proof by interval connected + optimal interval decomposition + number of intervals in optimal interval decomposition $>1$ is disconnected

connected-imply-closure-connected_(tag) $𝑆$ is a connected set ==> $\overline{𝑆}$ is a connected set

Proof

$\overline{𝑆}$ is not a connected set ==> $𝑆$ is not a connected set

connected-componet_(tag) Connected component decomposition := Limit point set decomposition's limit $𝑋=\text{ lim }{⨆}_{𝑖\in 𝐼}{𝐴}_{𝑖}$, such that each limit point set ${𝐴}_{𝑖}$ cannot be further decomposed i.e. connected

It is indeed the unique limit in the sense of #link(<net>)[net]. The net comes from the decomposition of $𝑋$ into two closed sets, which can be taken as a common refinement decomposition + closed sets are closed under finite intersection.

$𝑆$ is ${\text{T}}_{𝑆}$ connected or $𝑆$ cannot be ${\text{T}}_{𝑆}$ closed set decomposition and ${\text{T}}_{𝑋}$ has closed set decomposition $𝐴\bigsqcup {𝐴}^{\prime }$ ==> $\left(𝑆\subset 𝐴\right)\oplus \left(𝑆\subset {𝐴}^{\prime }\right)$

Proof The closed set decomposition of ${\text{T}}_{𝑆}$, $\left(𝑆\cap 𝐴\right)\bigsqcup \left(𝑆\cap {𝐴}^{\prime }\right)$, results in one of the sets being an empty set

$𝐴$ is a limit point connected set ==> $𝐴$ is in the only one limit point connected component of $𝑋$

Proof The points of $𝐴$ must be in $𝑋$ and therefore in some connected component.

==> Even if $𝑋$ is only decomposed into connected sets, it is already a connected component decomposition.

The union of connected sets ${𝑆}_{𝑖}$ with a common point $𝑥$, ${\bigcup }_{𝑖\in 𝐼}{𝑆}_{𝑖}$, is connected

Proof The connected sets containing $𝑥$ are all in the same connected component. This shows that ${\bigcup }_{𝑖\in 𝐼}{𝑆}_{𝑖}=𝑋$ has only one connected component, and is therefore connected.

A connected component is a maximal element of the $\subset$ #link(<maximal-linear-order>)[maximal linear order] of a connected set family.

The image of a continuous function transmits connectedness.

The inverse-image of a continuous function transmits disconnectedness as contrapositive

Proof Closed set decomposition $𝑌=𝐴\bigsqcup {𝐴}^{\prime }$ ==> Closed set decomposition $𝑋={𝑓}^{-1}\left(𝐴\right)\bigsqcup {𝑓}^{-1}\left({𝐴}^{\prime }\right)$

==> mean-value-theorem-continuous_(tag) Intermediate Value Theorem for Continuous Functions. The image $𝑓\left(𝑋\right)$ of a continuous function $𝑓:𝑋\to \mathrm{ℝ}$ is connected #link(<real-connected-is-interval>)[therefore] is an interval

If any two points in $𝑌$ are in some connected subset $𝑆$, then $𝑌$ is connected. Proof let $𝑌=𝐴\bigsqcup {𝐴}^{\prime }$ with $𝐴,{𝐴}^{\prime }$ closed, prove that $𝐴\vee {𝐴}^{\prime }=\varnothing$. Or $𝑌={\bigcup }_{𝑦\in 𝑌}𝑆\left({𝑦}_0,𝑦\right)$ and the union of connected sets $𝑆\left({𝑦}_0,𝑦\right)$ that have a common point ${𝑦}_0$ is connected

==> let $𝑋$ be connected. If any two points in $𝑌$ are in some connected image $𝑓\left(𝑋\right)$ of a continuous function, then $𝑌$ is connected

==> Path connected

product-topology-preserve-connected_(tag) #link(<product-topology>)[Product topology] preserves connectedness

Proof

let the connected component decomposition ${𝑋}_{𝑖}={\prod }_{𝑗\in 𝐽\left(𝑖\right)}{𝐴}_{𝑖,𝑗\left(𝑖\right)}$

All connected components of $\prod _{𝑖\in 𝐼}{𝑋}_{𝑖}$ are

$$\prod _{𝑖\in 𝐼}{𝐴}_{𝑖,𝑗\left(𝑖\right)}:𝑗\in \prod _{𝑖\in 𝐼}𝐽\left(𝑖\right)$$

Proof Using #link(<dependent-distributive>)[] ${\prod }_{𝑖\in 𝐼}{⨆}_{𝑗\in 𝐽}{𝐴}_{𝑖,𝑗\left(𝑖\right)}={⨆}_{𝑗\in 𝐽}{\prod }_{𝑖\in 𝐼}{𝐴}_{𝑖,𝑗\left(𝑖\right)}$ and the product being connected implies product connectedness, so ${\prod }_{𝑖\in 𝐼}{𝐴}_{𝑖,𝑗\left(𝑖\right)}$ is connected, thus it can no longer be decomposed

Define (how?) the topology or limit point of $𝑓\in 𝐶\left(𝑋\to 𝑌\right)$ (should be something compact open topology?)

homotopy_(tag) homotopy or limit point homotopy := $𝐶\left(𝑋\to 𝑌\right)$ is limit point connected

Example ${\mathrm{ℝ}}^{𝑛+1}\setminus 0$ is homotopic to ${\mathrm{𝕊}}^{𝑛}$

homotopy-class_(tag) := the connected component of $𝐶\left(𝑋\to 𝑌\right)$

Since composition preserves continuity, composition induces an operation on $𝐶\left(𝑋\to 𝑌\right)$. Prove whether it is well-defined. Sometimes it's invertible, making it a group operation