connected components in box topology Consider $R^{\omega}$ in the box topology. 
x & y are in the same component iff $x_i = y_i$ for infinitely many values.
I proved the --< implication, but I can't come up with the proof for the other direction. I was thinking to do it by contradiction that is suppose x & y are in the same components and $x_i \neq y_i$ for infinitely many values, but I couldn't proceed further. The problem that I am having is that in order to show this I have to show there doesn't exist any connected subspace of $R^{\omega}$ containing both x and y.
proof:
Suppose that x and y are in same component and $x_i = y_i$ for finitely many values of $i \in \mathbb{Z^{+}}$.
Let $\phi_i : R \rightarrow R$ be a homeomorphism that depend on i such that $\phi_i(x_i) = 0$ and $\phi_i(y_i) = i$ when $y_i \neq x_i$.
Consider $\Pi\ f_i = \tau\ : \ R^{\omega} \rightarrow R^{\omega}$ this is a homeomorphism in $R^{\omega}$ with the box topology (trivial to proof).
Now here is where I am having troubles;I can see intuitively that we will have connected component U and V that separate $\tau(x)$ and $\tau(y)$ but I can't prove that rigorously if someone could help in that final step that would be great. 
 A: This was originally published as part of this note which covered all three Munkres topologies on $\mathbb{R}^\omega$ (based on the posting on topology atlas forums. For easier reference on this site I re-use the most complicated part on the box topology:
Let $X = \mathbb{R}^\omega$ be a countable product of copies of $\mathbb{R}$., 
and we give $X$ the box-topology.
Define a relation $\sim$ on $X$ as follows:
$x \sim y$ iff the sequence $x_n - y_n$ is $0$ from a certain index 
onwards (or equivalently if the set of $\{n: x_n \ne y_n\}$ is finite). 
This is an equivalence relation:
$x \sim x$ because then we have a $0$-sequence, symmetry is evident, as 
$-0 = 0$, and if $N_1$ is an index from which $x_n - y_n = 0$, and $N_2$ a 
similar one for $y_n - z_n$, then $\max\{N_1,N_2\}$ works for $x_n$ and 
$z_n$.
I'll show first that the classes $[x]$ are path-connected.
So fix $x$. 
Then define for each finite subset $I$ of $\omega$ the set 
$X(I) = \{(y_n) : y_n = x_n \text{ for all $n$ in $\omega \setminus I$}\}$.
This is homeomorphic to $\prod_{n \in I} \mathbb{R}$ by the obvious map
(and by noting that on a finite(!) product the box topology and the normal 
product topology agree, and a finite (normal) product of copies of 
$\mathbb{R}$ is path-connected). 
Also for all such $I$, $x$ is in $X(I)$.
And $[x] = \bigcup\{X(I) : I \subset N, \text{ $I$ finite}\}$, by 
definition.
And this union is path-connected because we can always connect
two points in different $X(I)$'s via their common point $x$.
Claim (to finish): if $p$ is a point not in $[x]$, so it differs with 
$x_n$ for infinitely many $n$, then there is a subset $Y$ such that 
$p \in Y$ and $Y$ clopen (i.e. closed and open) and $x \notin Y$.
This shows that no connected subset can contain both $x$ and $y$, so 
$[x]$ is a maximal connected subset and hence a (path-)component of $X$.
To this end: 
let $U_{k,n}$ be open subsets of $\mathbb{R}$ (in the $n$-th component 
space) such that


*

*$x_n \in U_{k,n}$ for all $k$ and $n$ in $\omega$,

*$p_n \not\in U_{k,n}$ unless $p_n = x_n$,

*$\operatorname{cl} U_{{k+1},n} \subset U_{k,n}$ for all $k$ and $n$ in $\omega$.


It is obvious such sets can be chosen.
Define 
$$
Y = \{(y_n) : \text{there is a $k$ such that for infinitely many $n$ we 
have $y_n \not\in U_{k,n}$}\}.
$$
This $Y$ does not contain $x$, as then for all $k$,$n$ we have 
$x \in U_{k,n}$.
It does contain $p$, as $p$ differs on infinitely many places from $x$, so 
for every $k$ there are infinitely many $n$ such that 
$p_n \not\in U_{k,n}$, by (2) above.
It remains to show that $Y$ is closed and open.
$Y$ is open: let $z$ be in $Y$. 
Let $k$ be given as in the definition of $Y$. 
So there are infinitely many indices $n$ (say $n$ in $N'$) such that $z_n$ 
is not in $U_{k,n}$.
In particular, such $z_n$ are in the complement of $U_{k+1,n}$.
Then the box-topology basic open subset $\prod_{n \in \omega} O_n$ where 
$O_n$ is the complement $U_{k+1,n}$, if $n$ in $N'$
and $O_n$ is $\mathbb{R}$ otherwise, is contained in $Y$ and contains $z$.
(It is contained in $Y$ because $k+1$ works in the definition for $Y$ for 
all points of this open set.)
Y is closed: 
let $z \not\in Y$. 
So for all $k$ there are only finitely many $n$ such that 
$z_n \not\in U_{k,n}$. 
Call these finite exception sets $I_k$.
These sets are increasing: if $n$ is in $I_m$ then also in $I_{m+1}$.
If $n$ is in neither of the $I_k$, $z_n$ must be in the intersection (over 
$k$) of the $U_{k,n}$. 
Let $f$ be some strictly increasing map from 
$\omega \setminus \bigcup I_k$ into $\omega$ (if this set 
is non-empty).
Define the following open set 
$O = \prod_{n \in \omega} O_n$ in the box topology:
if $n$ is in $I_{k+1} \setminus I_k$, for some $k \ge 0$, let $O_n$ be 
$U_{k,n} \setminus \operatorname{cl} U_{k+1,n}$,
for $n$ in $I_0$, let $O_n$ be $\mathbb{R}$, and for $n$ in 
$\omega \setminus U_{k,n}$.
I claim that $O$ contains $z$ (this is quite clear), and that $O$ is 
disjoint from $Y$.
Why is this so? 
Let $w$ be in $O$. 
Pick $k$ in $\omega$. 
If $w_n$ is not in $U_{k,n}$, then we want to show there can be only 
finitely many of these $n$. 
Consider the cases for $n$:
$n$ can be in $I_{k+1} \setminus I_k$ only if $i+1 \le k$, so there are 
only finitely many $k$ that can apply.
And the union of these is finite. 
If $n$ is in $I_0$, who cares? 
Only finitely many $n$ are in $I_0$.
And finally, if $n$ is in 
$\omega \setminus \bigcup I_k$, this can only be a problem if this 
is set is infinite,
but then $f$ grows larger than $k$ from some index $N$ onwards,
and then $w_n$ is in $U_{f(n),n} \subset U_{k,n}$ (remember that the 
$U$'s decrease in $k$)
for $n \ge N$. 
So in $\omega \setminus \bigcup I_k$ there can also only be finitely 
many $n$ with $w_n \not\in U_{k,n}$.
So this shows that for every $k$, all but finitely many $n$ have 
$w_n \in U_{k,n}$.
So $w$ is not in $Y$, and $O$ is disjoint from $Y$.
So $Y$ is closed.
