Let S be a subset of $\mathbb{C}$. If $z,w \in S$, define $z ∼ w$ if and only if there is a path from z to w. Show that ∼ is an equivalence relation.
Definition of path: A path in $\mathbb{C}$ is a continuous function $\gamma:[a,b] \longrightarrow \mathbb{C}$, $\gamma(a)$ = initial point, $\gamma(b) =$ end point.
From my understanding, I need to show 3 components: reflective, symmetric, and transitive.
For Reflective, I have: Suppose $z_1 = x_1 +iy_1 \in \mathbb{C},$ then $x_1 +iy_1 - (x_1 +iy_1) = 0$ is a path
Symmetric: Suppose $z_1=x_1 +iy_1 \in \mathbb{C}$ and $w_1 = x_2+i+y_2 \in \mathbb{C}$ and $z_1Rz_2$, then there is a path from z to w. Since $\gamma:[x_2,y_2]\longrightarrow \mathbb{C}$, there is a path from w to z.
Transitive: Suppose $z,w \in \mathbb{C}$ and z~w and w~y, then there is a path from z to w and from w to $y=x_3+iy_3$. Thus $\gamma:[x_1,y_3]\longrightarrow \mathbb{C}$ so there is a path from z to y.
Am I going about these the right way?
As I said in my comment, $[a, b]$ is an interval in $\mathbb{R}$. For reflexivity, let $x \in S$ and let $\gamma:[0, 1] \rightarrow \mathbb{C}$ be the constant map given by $\gamma(t) = x$ for all $t \in [0,1]$. Then $\gamma$ is clearly continuous, and $\gamma(0) = \gamma(1) = x$. So that's a path from $x$ to $x$.
Now suppose there is a path $\gamma : [a, b] \rightarrow \mathbb{C}$ from $x$ to $y$, i.e. $\gamma(a) = x$ and $\gamma(b) = y$. Then let $\tilde{\gamma} : [a, b] \rightarrow \mathbb{C}$ be defined by $$\tilde{\gamma}(t) = \gamma(b+a-t)$$ Then $\tilde{\gamma}$ is continuous because $\gamma$ is, and $\tilde{\gamma}(a) = \gamma(b+a-a) = \gamma(b) = y$, and $\tilde{\gamma}(b) = \gamma(b+a-b) = \gamma(a) = x$. So $\tilde{\gamma}$ is a path from $y$ to $x$.
For transitivity, suppose $\gamma:[0, 1] \rightarrow \mathbb{C}$ is a path from $x$ to $y$ and $\theta: [0, 1] \rightarrow \mathbb{C}$ is a path from $y$ to $z$. Define $\delta: [0, 1] \rightarrow \mathbb{C}$ by $$\delta(t) = \gamma(2t), 0 \leq t \leq \frac{1}{2}$$ $$\delta(t) = \theta(2t-1), \frac{1}{2} < t \leq 1 $$ Note that $\delta$ is continuous at $t = \frac{1}{2}$. This is a path from $x$ to $z$.