Path connectedness as an equivalence relation

Question

Let $(X,\tau)$ be a topological space, and let $\sim$ be an equivalence relation such that $x\sim y$ iff there's a path connecting $x,y$.

1. $x\sim x$ because every singleton is connected with the constant path.
2. $x \sim y\rightarrow y \sim x$ because if $\gamma:[a,b]\rightarrow(X,\tau)$ is a path from $x$ to $y$, then $\gamma\circ\phi:[a,b]\rightarrow(X,\tau),\ \phi(t)=a+b-x$ is a path from $y$ to $x$.
3. As for transitiveness, I'm a bit stuck. if $\alpha:[a,b]\rightarrow(X,\tau),\ \beta':[n,m]\rightarrow(X,\tau)$ are pathx from x to y and from y to z, we can adjust $\beta'$ to $\beta:[b,c]\rightarrow (X,\tau)$, then we can concatenate the two functions to $\gamma:[a,c]\rightarrow (X,\tau),\ \gamma(x)=\begin{cases}\alpha(x),\ x\in[a,b]\\\beta(x),\ x\in (b,c]\end{cases}$
   the source for any open set $A$ in $\tau$ is the union of sources $\alpha^{-1}(A)\cup\beta^{-1}(A)$. If the two are disjoint and don't include $b$, then they're a union of open sets in $[a,c]$ as well, hence an open set. If they aren't disjoint, then the intersection is $b$, hence the union is a union of disjoint open sets and an open interval on $[a,c]$ (that includes $b$), hence an open set. However, if the $\alpha^{-1}(A)\cap\beta^{-1}(A)=\emptyset$ and let's say WLOG $b\in \alpha^{-1}(A)$, then the the union isn't an open set in $[a,c]$ (since b in the union has no neighborhood).
   I guess the last scenario then isn't possible, why though?

Answer 1

I take it you are trying to prove that $\gamma$ is continuous. Are you familiar with a theorem sometimes called the pasting lemma?

Suppose $\tilde{X}$ is a topological space and $E,F \subset \tilde{X}$ closed sets whose union $E \cup F = \tilde{X}$. Further suppose that $f:E \to X$ and $g:F \to X$ are maps of these respective closed sets into the same target space which agree on the intersection of their domains; that is, $f(x)=g(x)$ for any $x \in E \cap F$. We may then define

$h:\tilde{X} \to X \, ; \, x \mapsto \begin{cases} f(x), & x \in E \\ g(x), & x \in F \end{cases}$.

The pasting lemma says that this map $h$ is continuous whenever $f,g$ are continuous.

Proof: Suppose $C \subset X$ is closed. Then $h^{-1}(C) =( f^{-1}(C) \cap E) \cup (g^{-1}(C) \cap F)$ is closed in $\tilde{X}$ by continuity of $f,g$. Thus $h$ is continuous.

In your situation $\tilde{X} = [a,c] = [a,b] \cup [b,c]$ and you have continuous paths $\alpha:[a,b] \to X$ and $\beta: [b,c] \to X$ that agree at the intersection $[a,b] \cap [b,c] = \{b \}$ of their domains, simply because $\alpha(b) = y = \beta(b)$. By the pasting lemma,

$\gamma:[a,c] \to X \, ; \, t \mapsto \begin{cases} \alpha(t), & t \in [a,b] \\ \beta(t), & t \in [b,c] \end{cases}$

is a continuous function.