How to prove ~ is an equivalence relation In a metric space $M$, declare $x \sim y$ to mean that there is a continuous function $\gamma : [0, 1] \rightarrow M$ such that $\gamma(0) = x$ and $\gamma(1) = y$. Prove that $\sim$ is an equivalence relation.
Clearly I must show that $\sim$ is reflexive, symmetric and transitive. Though I am not sure how to do so in this case.
My thought for reflexive is: Let $x\in M$. Then I provide a function $\gamma : [0, 1] \rightarrow M$ such that $\gamma(0)=x$ and $\gamma(1)=x$. So I let $\gamma(x)=x$. Thus $\sim$ is reflexive since $\gamma(x)=x$ is continuous right?
Then for symmetric: I let $x,y\in M$ and assume $x\sim y$. So I know that there exists a continuous function $\gamma : [0, 1] \rightarrow M$ such that $\gamma(0) = x$ and $\gamma(1) = y$. Then I believe that I need to show that there exists a continuous function $\alpha$ such that $\alpha(0) = y$ and $\alpha(1) = x$. Is this correct and how can I do so?
I'm sure once I figure this part our transitivity will follow easily.
 A: For reflexivity, you have the right idea, except you don't want to use $x$ as the input of $\gamma$ and the output ($M$ may not contain the interval $[0,1]$). Instead you should write something like $\gamma(t)=x$ (and yes, this is continuous since it is a constant function).
For symmetry, take a look at @BrianO's suggestion. The idea is that you want to reverse the roles of 0 and 1, so when you input $0$ into $\alpha$, it's the same as inputting $1-0=1$ into $\gamma$, and when you input $1$ into $\alpha$, it's the same as inputting $1-1=0$ into $\gamma$.
For transitivity, it may be a little tricky to write down, but the idea is this: if $x\sim y$ and $y \sim z$, there are continuous functions $\gamma:[0,1] \to M$ and $\beta:[0,1]\to M$ such that $\gamma(0)=x$, $\gamma(1)=y$, $\beta(0)=y$, and $\beta(1)=z$. Think about "shifting" $\beta$ over so it is defined on $[1,2]$; that is, define $\hat\beta:[1,2]\to M$ such that so that $\hat\beta(1)=y$ and $\hat\beta(2)=z$ (how should $\hat\beta$ be defined in terms of $\beta$?). Now think about "gluing" $\gamma$ and $\hat\beta$ at the point $1$ to get a continuous function $\alpha:[0,2]\to M$ with $\alpha(0)=x$ and $\alpha(2)=z$ (why is $\alpha$ continuous?). Finally, use $\alpha$ to define a continuous function on $[0,1]$ with the properties you want.
