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.


  1. Is my partial solution correct?
  2. How to figure out the other parts of the problem?
  • 1
    $\begingroup$ at.yorku.ca/p/a/c/a/17.pdf has my write-up of this problem. $\endgroup$ Nov 15, 2015 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, 2017 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, 2017 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. $\endgroup$ Nov 16, 2017 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, 2017 at 17:32

1 Answer 1


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.

  • 1
    $\begingroup$ 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. $\endgroup$ Feb 8, 2015 at 22:56
  • 1
    $\begingroup$ @Joseph: You’re absolutely right; I don’t know what I was thinking. It should be okay now, however. $\endgroup$ Feb 9, 2015 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. $\endgroup$ May 9, 2015 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. $\endgroup$ May 9, 2015 at 19:51
  • 1
    $\begingroup$ @BrianM.Scott Thank you for a wonderful answer. However, in box topology section, mind if I ask you a clarification? In your proof, you claim "$U$ is a clopen set separating $x$ and $y$, and the component of $x$ must therefore be a subset of $F(x)$." Here, are you implying that $F(x) = \mathbb{R}^\omega \setminus U$? If so, then $\mathbb{R}\setminus F(x) \subset U$ is clear, but $\mathbb{R}\setminus F(x) \supset U$ is not straightforward for me. $\endgroup$
    – James C
    Jul 24, 2020 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.