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:
- Is my partial solution correct?
- How to figure out the other parts of the problem?
Best Answer
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.