Definition of path-connected spaces 
This definition is from Munkres' Topology. I have a few questions about this and the proof for the fact that every path-connected space is connected.
(1) In the definition of path-connectedness, do the map $f$ and the closed interval $[a,b]$ depend on the points $x$ and $y$? In other words, can I rephrase the definition as follows?
A space $X$ is path connected iff for every pair of points $x$ and $y$ in $X$, there exist a closed interval $[a,b]\subset\mathbb{R}$ and a continuous map $f:[a,b]\to X$, both of which depend on the points $x$ and $y$, such that $f(a)=x$ and  $f(b)=y$.
(2) I have seen other definition of path where the domain is fixed to be $[0,1]$. Is there any advantage of using an arbitrary closed interval than $[0,1]$ or are they equivalent?
(3) In proving the statement that every path-connected space is connected, what does Munkres mean by "let $f:[a,b]\to X$ be any path in $X$"? Does he mean "chose any two points $x$ and $y$ in X. Then, there exists a path $f:[a,b]\to X$ such that $f(a)=x$ and $f(b)=y$"?
I would appreciate any help.
 A: In (1) your wording is fine, except that the phrase "both of which depend on $x,y$" is irrelevant. Think about the logical structure of that universally quantified statement:


*

*For all $x,y \in X$, $P(x,y)$


The statement $P(x,y)$ is 


*

*there exists a closed interval $[a,b] \in \mathbb R$ and a continuous map $f : [a,b] \to \mathbb R$ such that $f(a)=x$ and $f(b)=y$ 


To say that $[a,b]$ and $f$ depend on $x$ and $y$ is implied in the logical structure of the statement: the nature of the dependence is written out in the portion of the statement $P(x,y)$ that follows the existential quantifier, namely 


*

*$f(a)=x$ and $f(b)=y$


You've seen this kind of thing before, I'm sure. For example, consider the definition of continuity of a function $f : \mathbb R \to \mathbb R$ at a point $a \in \mathbb R$, which has the form "for all $\epsilon > 0$, $P(\epsilon)$". The statement $P(\epsilon)$ is "there exists $\delta > 0$ such that if $|x-a|<\delta$ then $|f(x)-f(a)| < \epsilon$". You might ask the same question, namely whether $\delta$ depends on $\epsilon$. The answer is that this dependence is written out in the portion of the statement $P(\epsilon)$ that follows the existential quantifier, namely "if $|x-a|<\delta$ then $|f(x)-f(a)| < \epsilon$".
Look for other mathematical statements of the form "for all this there exists that such that something", and you'll see the same logical meaning.
A: (1,2) The domain is irrelevant, because you can always reparametrize a path via a linear function to get an equivalent path with another domain. For example, given a path $f:[a,b] \to X$, the path $f\circ g:[0,1] \to X$, with $g(t) = (1-t)a + tb$, traces out the same path in $X$ as $f$.
(3) No, he literally means "let $f:[a,b] \to X$ be any path in $X$", i.e., "let $f:[a,b] \to X$ be any continuous function". Then, you can take $x = f(a)$ and $y = f(b)$ to be the endpoints if you want, but they're irrelevant to his argument.
A: As to (1),(2): yes, the path depends on $x$ and $y$, which is obvious as it must obey $f(a) = x$ and $f(b) = y$. The domain is actually irrelevant, as long as it is a closed bounded interval like $[a,b]$. It is common to fix it at $[0,1]$, and if there is a path $f: [a,b] \to X$ from $x$ to $y$, then $\phi(t) = (b-a)t + a$ maps $[0,1]$ bijectively and continuously to $[a,b]$, and then $f' := f \circ \phi: [0,1] \to X$ is continuous as well and satisfies $f'(0) = f(\phi(0)) = f(a) = x$ and $f'(1) = f(\phi(1)) = f(b) = y$. So we could just as well define the notion of path-connectedness with $[0,1]$ as a fixed domain. (This is actually usual when discussing homotopies of paths later).
So we could define: $X$ is path-connected when 
$$\forall x,y \in X: \exists f:[0,1]\to X: f \text{ continuous and } f(0) = x, f(1) = y$$
where $f$ does depend on $x,y$ (they are quantified over first).
As to (3), the final bit of the argument really is (spelled out): we now know that a path always has both endpoints in either $A$ or $B$. But there is at least one point of $A$, say $a_0$ and at least one point in $B$, say $b_0$. By path-connectedness of $X$ (here we use it for the first time!) there is a path $f: [a,b] \to X$ from $a_0$ to $b_0$. But this contradicts our previous fact because the endpoints lie in different parts, so contradiction: there cannot be a disconnection of $X$ and $X$ is connected. But Munkres expects you to fill in this last bit yourself.
