Connected space/Proofs

Every path-connected space is connected

Let "S" be path-connected and suppose, for contradiction, that "S" is not connected. Then $S\; =\; A\; cup\; B$ for nonempty disjoint open sets "A" and "B". Let $x\; in\; A,\; y\; in\; B$. Since "S" is path-connected, there exists a continuous path $gamma:\; [0,\; 1]\; o\; S$ with $gamma(0)\; =\; x$ and $gamma(1)\; =\; y$.

Since "f" is continuous, $ilde\{A\}\; =\; f^\{-1\}(A)$ and $ilde\{B\}\; =\; f^\{-1\}(B)$ are open subsets of [0, 1] . Moreover $ilde\{A\}$ and $ilde\{B\}$ are nonempty since they contain 0 and 1, respectively. They are disjoint since "A" and "B" are disjoint, and $ilde\{A\}\; cup\; ilde\{B\}\; =\; [0,\; 1]$, so [0, 1] is not connected, a contradiction.

Therefore "S" is connected.

A locally path-connected space is path-connected if and only if it is connected

A path-connected space is always connected. We therefore focus on the converse.

Let "S" be connected and locally path-connected. We for points "a" and "b" in "S", we denote by the relation "a" ~ "b" that there exists a path between "a" and "b".

Lemma 1: ~ is an equivalence relation

We check:

#Reflexivity: "a" ~ "a", as evidenced by the trivial path $gamma(t)\; =\; a$.
#Symmetry: Suppose "a" ~ "b", and let $gamma(t)$ be the associated path between "a" and "b". Then $phi(t)\; =\; gamma(1-t)$ defines a continuous path from "b" to "a", and thus "b" ~ "a".
#Transitivity: Suppose "a" ~ "b", with associated path $gamma(t)$, and "b" ~ "c", with associated path $phi(t)$. Then let

$psi(t)$ is then a continuous path from "a" to "c", so "a" ~ "c".

Lemma 2: For a point "a" in "S", the equivalence class ["a"] is open

Let $p\; in\; [a]$. Then since "S" is locally path-connected, there is a neighborhood "U" of "p" so that, for every $q\; in\; U$, there is a path from "p" to "q". But then ["q"] = ["p"] = ["a"] , so $U\; subset\; [a]$ and "p" is an interior point of ["a"] . Hence ["a"] is open.

Lemma 3: ["a"] is closed

Let "C" denote the complement of ["a"] . Then

The union of open sets is open, and each term of the union is open by Lemma 1, so "C" is open, and hence ["a"] is closed.

Proof of theorem

Let "x" and "y" be two points of "S". Then by Lemmas 1 and 2, ["x"] is clopen, so since "S" is connected, ["x"] = "S". Hence $y\; in\; [x]$, and there exists a path between "x" and "y", and "S" is path-connected.