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I am working on an exercise problem about components and path components of $\mathbb{R}^{\omega}$. Specifically,

Exercise about components and path components:
1. What are the components and path components of $\mathbb{R}^{\omega}$ in the product topology?
2. What are the components and path components of $\mathbb{R}^{\omega}$ in the uniform topology?
3. What are the components and path components of $\mathbb{R}^{\omega}$ in the box topology?

I can only handle with only parts of the problem (about components):

My partial solution:
1. $\mathbb{R}^{\omega}$ in the product topology is connected, so its only component is $\mathbb{R}^{\omega}$.
2. $\mathbb{R}^{\omega}$ in the uniform topology is not connected (see here). There are two components: $A$ consisting of all bounded sequences of real numbers and $B$ of all unbounded sequences. [EDIT: I realized that the answer is wrong: $A$ and $B$ constitute a separation of $\mathbb{R}^{\omega}$ in the uniform topology. However, this does not imply that $A$ and $B$ are two components of it. So, I have no idea of this problem.]
3. No idea.

Therefore:

  1. Is my partial solution correct?
  2. How to figure out the other parts of the problem?
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    $\begingroup$ at.yorku.ca/p/a/c/a/17.pdf has my write-up of this problem. $\endgroup$ – Henno Brandsma Nov 15 '15 at 9:50
  • $\begingroup$ @HennoBrandsma I am having trouble understanding the proof you linked. First, how does prove that $Z$ is clopen? I tried writing it as the union and intersection of preimages under some continuous map, but I had no luck. Second, why does $Z$ being clopen, and containing $x$ but not $y$, entail that $x$ and $y$ cannot in the same component? If I am not mistaken, the only way for this to be is if there is no connected set that contains both $x$ and $y$; but I don't see how $Z$ having the aforementioned properties implies this. $\endgroup$ – user193319 Nov 16 '17 at 15:10
  • $\begingroup$ @HennoBrandsma Also, in your definition of the set $Z$, why do you write "...such that for all p in N"? Sure there are infinitely values at which x(a_p) - y(a_p) is not zero, but that doesn't mean there are none, which means that you could possible divide by $0$. $\endgroup$ – user193319 Nov 16 '17 at 15:14
  • $\begingroup$ @user193319 I define a set $Y$ that is clopen. This is shown by proving $Y$ is open and its complement as well. I don’t see what $Z$ you mean. And dividing by $0$ I don’t see at all. Please refer to exact locations for questions. $\endgroup$ – Henno Brandsma Nov 16 '17 at 16:54
  • $\begingroup$ @HennoBrandsma Whoops! I had two different, but related, MSE posts open at once. I commented on this one accidently, and I can't seem to find the other MSE post. At any rate, in this other MSE post you made a comment in which you provided this link: at.yorku.ca/cgi-bin/… $\endgroup$ – user193319 Nov 16 '17 at 17:32
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Your partial solution is correct as far as it goes. Path connectedness, like connectedness, is productive (see here or here), so $\Bbb R^\omega$ is also its only component in the product topology.


Identifying the components of $\Bbb R^\omega$ in the uniform topology is a little tricky. It’s easiest to start with an arbitrary $x=\langle x_n:n\in\omega\rangle\in\Bbb R^\omega$ and describe the component $C(x)$ containing $x$. In fact, it’s easiest to start with $C(z)$, where $z=\langle 0,0,0,\ldots\rangle$ is the point with all coordinates $0$. Let $B$ be the set of bounded sequences in $\Bbb R^\omega$; you already know that $B$ is clopen in the uniform topology, and I claim that it’s also connected and hence that $C(z)=B$.

Let $x=\langle x_n:n\in\omega\rangle\in B\setminus\{z\}$. Define

$$f:[0,1]\to\Bbb R^\omega:t\mapsto\langle tx_n:n\in\omega\rangle\;,$$

and note that $f(0)=z$ and $f(1)=x$. Let $t\in[0,1]$ and $\epsilon>0$ be arbitrary, and let $N$ be the $\epsilon$-ball centred at $f(t)$:

$$N=\{\langle y_n:n\in\omega\rangle\in\Bbb R^\omega:|y_n-tx_n|<\epsilon\text{ for all }n\in\omega\}\;;$$

what condition on $s\in[0,1]$ will ensure that $f(s)\in N$, i.e., that $|sx_n-tx_n|<\epsilon$ for all $n\in\omega$? Clearly we want to have $|s-t|<\frac{\epsilon}{|x_n|}$ for all $n\in\omega$. Is this possible? Yes, because $x\in B$, and therefore $\{|x_n|:n\in\omega\}$ is bounded. Let $\|x\|=\sup_n|x_n|$, and let $\delta=\frac{\epsilon}{\|x\|}$; then $f(s)\in N$ whenever $|s-t|<\delta$, and $f$ is therefore continuous. Thus, $B$ is even path connected and is both the component and the path component of $z$.

Now let $x\in\Bbb R^\omega$ be arbitrary; then $C(x)=B+x$, where as usual $B+x=\{y+x:y\in B\}$, and $C(x)$ is also the path component of $x$. To see this, simply note that the map $$f_x:\Bbb R^\omega\to\Bbb R^\omega:y\mapsto y+x$$ is a homeomorphism, that $x=f_x(z)$, and that $B+x=f_x[B]$. It’s not hard to see that we can also describe $C(x)$ as the set of $y\in\Bbb R^\omega$ such that $y-x$ is bounded.


Corrected Version (8 February 2015):

Finally, we consider $\Bbb R^\omega$ with the box topology. This is a significantly harder problem. For $x\in\Bbb R^\omega$ let $$F(x)=\big\{y\in\Bbb R^\omega:\{n\in\omega:x_n\ne y_n\}\text{ is finite}\big\}\;;$$ I’ll show first that $F(x)$ contains the component of $x$.

Suppose that $y\in\Bbb R^\omega\setminus F(x)$. For $n\in\omega$ let $\epsilon_n=|x_n-y_n|$. If $\epsilon_n>0$ and $k\in\omega$ let

$$B_n(k)=\left(x_n-\frac{\epsilon_n}{2^k},x_n+\frac{\epsilon_n}{2^k}\right)\subseteq\Bbb R\;,$$

and note that $y_n\notin B_n(k)$ for any $k\in\omega$. For each $k\in\omega$ let

$$U_k=\big\{z\in\Bbb R^\omega:\{n\in\omega:z_n\notin B_n(k)\}\text{ is infinite}\big\}\;,$$

and let $U=\bigcup_{k\in\omega}U_k$. Clearly $y\in U$ and $x\notin U$.

  • Let $z\in U$; then $z\in U_k$ for some $k\in\omega$. Let $M=\{n\in\omega:z_n\notin B_n(k)\}$. Then $$z_n\notin B_n(k)\supseteq\operatorname{cl}_{\Bbb R}B_n(k+1)$$ for each $n\in M$. For $n\in M$ let $G_n=\Bbb R\setminus\operatorname{cl}_{\Bbb R}B_n(k+1)$, and for $n\in\omega\setminus M$ let $G_n=\Bbb R$; then $G=\prod_{n\in\omega}G_n$ is open, and $z\in G\subseteq U_{k+1}\subseteq U$, so $U$ is open.

  • Now suppose that $z\in\Bbb R^\omega\setminus U$; then $\{n\in\omega:z_n\notin B_n(k)\}$ is finite for each $k\in\omega$. In other words, for each $k\in\omega$ there is an $m_k\in\omega$ such that $z_n\in B_n(k)$ for all $n\ge m_k$. Clearly we may assume that $m_k<m_{k+1}$ for each $k\in\omega$. For $m_k\le n<m_{k+1}$ let $G_n=B_n(k)$, and for $n<m_0$ let $G_n=\Bbb R$. Then $G=\prod_{n\in\omega}G_n$ is open, and $z\in G\subseteq\Bbb R^\omega\setminus U$, so $U$ is closed.

Thus, $U$ is a clopen set separating $x$ and $y$, and the component of $x$ must therefore be a subset of $F(x)$.

In fact $F(x)$ is the component and path component of $x$. To see this, suppose that $y\in F(x)$, and let $D=\{n\in\omega:x_n\ne y_n\}$, so that $D$ is finite. Let

$$Y=\{z\in\Bbb R^\omega:z_n=x_n\text{ for all }n\in\omega\setminus D\}\;;$$

then $Y$ is homeomorphic to $\Bbb R^{|D|}$. $\Bbb R^{|D|}$ is path connected, and a path in $\Bbb R^{|D|}$ transfers easily to a path in $Y$ and hence in $\Bbb R^\omega$ via the homeomorphism.

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  • $\begingroup$ The box topology on $\mathbb{R}^{\omega}$ is definitely not zero-dimensional. If $(x_{n})_{n}$ and $(y_{n})_{n}$ differ by only finitely many coordinates, then $(x_{n})_{n}$ and $(y_{n})_{n}$ belong to the same path-component. $\endgroup$ – Joseph Van Name Feb 8 '15 at 22:56
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    $\begingroup$ @Joseph: You’re absolutely right; I don’t know what I was thinking. It should be okay now, however. $\endgroup$ – Brian M. Scott Feb 9 '15 at 0:53
  • $\begingroup$ Well, @Brian M. Scott, all you've shown is that every point of the set $B$ of bounded sequences can be joined to the point $z = (0, 0, 0, \ldots)$ by a path. How can you then conclude that $B$ is path connected? And how can you go on to state that $B$ is the path component and the comonent of $z$, even if $B$ is path connected? I mean you shold have also shown that no points outside of $B$ can be joined to $z$ by a paht. $\endgroup$ – Saaqib Mahmood May 9 '15 at 11:27
  • $\begingroup$ @Saaqib: It’s trivial to show that if each point of $B$ is connected by a path to $z$, then $B$ is path connected: given points $p,q\in B$, take a path from $p$ to $z$ followed by one from $z$ to $q$. We already knew that the component and path component of $z$ had to be contained in $B$, simply because $B$ is clopen. $\endgroup$ – Brian M. Scott May 9 '15 at 19:51

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