Given point $x$ and $y$ of the space $X$, a path in $X$ from $x$ to $y$ is a continuous map $f:[a,b]\to X$ of one closed interval in the real line into $X$, such that $f(x)=a$ and $f(y)=b$. A space $X$ is said to be path connected if every pair of points of $X$ can be joined by a path in $X$.
Why do you we use $[a,b]$ interval? Isn’t $[0,1]$ interval is more explicit to use? and word “joined” in definition of path connected mean? Is it $\exists$ path from $x$ to $y$ and $\exists$ path from $y$ to $x$?
Path connected $\implies$ connected
Approach(1): Munkres’ proof. By lemma 24.1 corollary, $[0,1]$ is connected. By theorem 23.5, $f([0,1])$ is connected. By lemma 23.2, either $f([0,1])\subseteq A$ or $f([0,1]) \subseteq B$. If $f([0,1]) \subseteq A$, then $f(1)=b\in A$. Which is a contradiction, because $b\in B$. If $f([0,1]) \subseteq B$, then $f(0)=a\in B$. Which is a contradiction, because $a\in A$. Thus $\nexists$ separation of $X$. Hence $X$ is connected.
Approach(2): Let $A$ and $B$ form separation of $X$. Since $A,B\neq \emptyset$, $\exists a\in A$ and $\exists b\in B$. Since $X$ is path connected and $a,b\in X$, $\exists f:[0,1]=I \to X$ such that $f$ is continuous, $f(0)=a$ and $f(1)=b$. So $0\in f^{-1}(A)$ and $1\in f^{-1}(B)$. Thus $f^{-1}(A), f^{-1}(B)\neq \emptyset$. $f^{-1} (A)\cap f^{-1}(B)=\emptyset$. $f^{-1} (A)\cup f^{-1}(B)=[0,1]=I$. Since $f$ is continuous, $f^{-1}(A),f^{-1}(B)\in \mathcal{T}_I$. Thus $f^{-1}(A)$ and $f^{-1}(B)$ form separation of $[0,1]$. By lemma 24.1 corollary, $[0,1]$ is connected. Thus we reach contradiction. Is this proof correct? In this approach we don’t make use lots of fact that we used in approach(1).
Approach(3): https://courses-archive.maths.ox.ac.uk/node/view_material/50743. Direct proof. Proposition 1.89, page number 19.
Approach(4): let $f:X \to \{0,1\}$ be a continuous map. We need to show $f$ is constant. Let $x,y\in X$. Since $X$ is path connected, $\exists p:[0,1] \to X$ such that $p$ is continuous and $p(0)=x$ and $p(1)=y$. By theorem 18.2(c), $f \circ p:[0,1]\to \{0,1\}$ is continuous. Since $[0,1]$ is connected, $f\circ p$ is constant by definition of connectedness. So $f\circ p(0)=f(x)=f\circ p(1)=f(y)$. Thus $f(x)=f(y)$, $\forall x,y\in X$. Hence $f$ is constant and $X$ is connected.