About Path-connected 
Let $a=(a_1,a_2,...,a_k)$ and $b=(b_1,b_2,...,b_k)$ be points in k-dimensional space $\mathbb{R}^k$. A path from $a$ to $b$ is a continuous function on the unit interval $[0,1]$ with values in $\mathbb{R}^k$, a function $X:[0,1]\rightarrow \mathbb{R}^k$, sending $t\rightsquigarrow X(t)=(x_1(t),...,x_k(t))$, such that $X(0)=a$ and $X(1)=b$. If $S$ is a subset of $\mathbb{R}^k$ and if $a$ and $b$ are in $S$, define $a\sim b$ if $a$ and $b$ can be joined by a path lying entirely in $S$.
$\textbf{(a)}$ Show that $\sim$ is an equivalence relation on $S$. Be careful to check that any paths you construct stay within the set $S$.
$\textbf{(b)}$ A subset $S$ is $\textit{path connected}$ if $a\sim b$ for any two points $a$ and $b$ in $S$. Show that every subset $S$ is partitioned into path-connected subsets with the property that two points in different subsets cannot be connected by a path in $S$.
$\textbf{(c)}$ Which of the following loci in $\mathbb{R}^2$ are path-connected: $\{x^2+y^2=1\},\{xy=0\},\{xy=1\}$

For $\textbf{(a)}$, I should test three properties: "transitive", "symmetric", "reflexive". The test of "symmetric" and "reflexive" are easy, and my work for "transitive" is as follows:
transitive:
$a\sim b :X_1(0)=a,X_1(1)=b$ ;
$b\sim c: X_2(0)=b,X_2(1)=c$ .
Then I construct a function $X_3(t)=(1-t)X_1(t) + tX_2(t)$ , and we have $X_3(0)=a,X_3(1)=c$.
But after working with $\textbf{(c)}$ , I find that $X_3$ , which I have constructed , may not stay within the set $S$. How should I construct it.
For $\textbf{(b)}$ , I think it follows from $\textbf{(a)}$ , Since an equivalence relation on a set $S$ determines a partition of $S$. Is that satisfied?
For $\textbf{(c)}$, I have constructed functions to prove the first two , but for the last, i.e. $\{xy=1\}$ , I know $a(-1,-1),b(1,1)$ could be a counterexample, but I don't know how to prove it.

Let $U:=\{(x,y)\in \mathbb{R}^2:x,y>0\}$ and $V:=\{(x,y)\in \mathbb{R}^2:x,y<0\}$. Then $U$ and $V$ form a separation of $\{(x,y)\in \mathbb{R}^2:xy=1\}$. Since $\{(x,y)\in \mathbb{R}^2:xy=1\}$ is not connected, it is not path-connected.
