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.
 A: You are correct that the function $X_{3}$ as you've constructed it doesn't solve this problem. What you need to do for $X_{3}$ to prove transitivity (and what I assume you essentially did for the set $\{(x,y) | xy = 0\}$) is construct a new path that follows first $X_{1}$ and then $X_{2}$:
$$ X_{3}(t) = \begin{cases} X_{1}(2t) & \textrm{ if } & 0 \leq t \leq 1/2 \\ X_{2}(2t-1) & \textrm{ if } & 1/2 \leq t \leq 1  \end{cases}.$$
You have to say a few words about continuity around $t=1/2$, but it's not so bad.
As for the last part, can you use the notion of Connected? I'm guessing that you don't yet have the fact "Path-connected implies Connected", which would be useful. But you can essentially make that argument directly. Assume you have a path $X(t)$ from $(1,1)$ to $(-1,-1)$. Focus on its components: $X(t) = (f(t),g(t))$. Then $f$ and $g$ are continuous real-valued functions. Apply the intermediate value theorem to show that that $f(t) = 0$ at some point $0 \leq t \leq 1$. Then at that point, $xy \neq 1$, and you've left the set $\{xy = 1\}$.
A: 
Show that $\sim$ is an equivalence relation on $S$. Be careful to check that any paths you construct stay within the set $S$.

Fix $a,b\in S$. The constant path $X(t)=a$ from $a$ to $a$ shows that $a\sim a$. If $a\sim b$ and $X$ is a path from $a$ to $b$ entirely in $S$, then $X'(t) = X(1-t)$ is a path (the "reverse" of $X$) from $b$ to $a$ entirely in $S$, so $b\sim a$.
With reflexivity and symmetry established, it remains to exhibit transitivity. If$$a\sim b,\text{ }b\sim c\text{ for some }a,\,b,\,c\in S,$$with $X$ taking $a$ to $b$ and $Y$ taking $b$ to $c$ (both $X$, $Y$ entirely in $S$), then concatenating (and scaling by $1/2$) $X$, $Y$ gives a valid path from $a$ to $c$. More explicitly,$$Z:[0,1]\to\mathbb{R}^k,\text{ }Z(t) = \begin{cases} X(2t) & \text{if }t \in [0,1/2] \\ Y(2t-1) & \text{if }t \in (1/2,1]\end{cases}$$ takes $a$ to $c$, stays in $S$ and is continuous on both intervals $[0,1/2]$ and $(1/2,1]$, and is continuous at $1/2$ as well, because $$\lim_{r\to1^-} X(r) = X(1) = b = Y(0) = \lim_{r\to0^+} Y(r).$$

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 $S$.

By definition, each equivalence class$$[a] = \{s\in S: s\sim a (\iff a\sim s)\}$$ of $\sim$ is path-connected. But for $a$, $b\in S$, the following three statements are equivalent:


*

*$a\sim b$;

*$[a] = [b]$;

*$[a]$, $[b]$ have nonempty intersection.


Symmetry and transitivity together yield $(1)\implies (2)$ and $(3)\implies (1)$; the non-emptiness of equivalence classes, from reflexivity, gives $(2)\implies (3)$.
To finish, just note that $(3)\implies(2)$ and $a\in [a]$ together show that $\sim$ partitions $S$ into path-connected components (equivalence classes), while $\neg(2)\implies\neg(1)$ proves that two points in distinct components cannot be connected by a path in $S$.
A: 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.
