# Logical Point in Topological Problem

1. May 4, 2017

### Bashyboy

1. The problem statement, all variables and given/known data
Consider $\mathbb{R}^\omega$ in the uniform topology. Show that $x$ and $y$ lie in the same component if and only if $x-y = (x_1-y_1,x_2-y_2,...)$ is a bounded sequence.

2. Relevant equations

The uniform topology is induced by the metric $p(x,y) := \sup_{i \in \mathbb{N}} d(x_i,y_i)$, where $d(x_i,y_i) = \min \{|x_i-x_j|,1\}$.

3. The attempt at a solution

My first question is, by bounded sequence does the author mean a bounded sequence in $\mathbb{R}$ and with respect to the metric on $\mathbb{R}$, which would just be the absolute value function?

My next question pertains to the hint. When the author say it suffices to show such-and-such, do the mean that the special case

$x \sim 0$ if and only if $x-0$ is a bounded sequence

implies the more general case

$x \sim y$ if and only if $x-y$ is bounded;

and therefore I need to prove the implication to show that it is an actual reduction, right?

2. May 4, 2017

### Dick

Yes, the usual metric in $\mathbb{R}$.

You didn't quote the hint part, but it's pretty easy to show those two are equivalent, isn't it? The metric is translation invariant.

Last edited: May 4, 2017
3. May 5, 2017

### Bashyboy

Okay. Perhaps you could critique this attempt at proving they are equivalent.

Let (1) refer to the special case, and (2) the more general case. First, note that $f_y : \mathbb{R}^\omega \to \mathbb{R}^\omega$ defined by $f_y(x) = x+y$ is a homeomorphism, whose inverse is $f^{-1}_y = f_{-y}$. First we will prove that (1) implies (2). Suppose that the (1) is true. Then $x-y = (x-y) - 0$ is, according to (1), bounded if and only if $(x-y) \sim 0$. This happens if and only if $x-y$ and $0$ are in the same connected subspace $C$. Since $f_y$ is continuous, the image $f_y(C)$ will also be connected in $\mathbb{R}^\omega$, and by definition $f_y(x-y) = x$ and $f_y(0) = y$ are in this connected subspace, i.e., $x \sim y$. And clearly if $x$ and $y$ are in the same connected subspace, we we can use the inverse function $f_{-y}$ to show $x-y$ and $0$ are in the same connected subspace Then $x \sim y$ if and only if x-y = (x-y) - 0 is bounded. But according to (1), this happens if and only if (x-y) \sim 0

Obviously proving (2) implies (1) starts almost the same way that the above proof ends, so we are finished.

4. May 5, 2017

### Bashyboy

Also, I am having trouble proving the claim "$x$ and $0$ lie in the same of $\mathbb{R}^\omega$ if and only if $x-0=x$ is bounded. I could use a hint on how to prove $\implies$ (forward) direction.

5. May 5, 2017

### Dick

Here's a notion to start from, let $B$ be the set of all bounded sequences. Can you show $B$ and $B^c$ are both open? Hint: consider open balls of radius 1 around points in each.

6. May 10, 2017

### Bashyboy

I am not sure how this would help. The only conclusion I could draw from $B$ being clopen is that $\mathbb{R}^\omega$ is not a connected space.

7. May 10, 2017

### Dick

No, not the only conclusion. It also tells you $B$ is a union of connected components. Your next job would be to show it consists of a single component (i.e. $B$ is connected).

8. May 15, 2017

### Bashyboy

Oh. Does your solution rely on clopen sets being the union of connected components? If so, I don't yet have this fact.

9. May 15, 2017

### Dick

It's sort of obvious, isn't it? The whole set is a union of its connected components - connected components are clopen. If you have doubts about this you might want to prove any missing parts as a exercise. Can it contain only part of a component?

Last edited: May 16, 2017
10. May 16, 2017

### Bashyboy

I am pretty certain this is false. Consider the rationals endowed with the standard (order) topology. Its components are singletons, which are closed but not open.

11. May 16, 2017

### Dick

Good catch. Sure. But in this case the components are open as well. Can you show that?

12. May 17, 2017

### Bashyboy

Everything I have read suggests that this is wrong, that connected components are not necessarily clopen. In the case of $\Bbb{Q}$, singletons are not open. Also, see JHance's comment here https://math.stackexchange.com/ques...e-components-of-the-set-of-irrational-numbers which has gone uncorrected for two years, suggesting, again, that connected components are not generally clopen.

13. May 17, 2017

### Dick

I acknowledged that that is correct. My point is that IN THIS CASE the components ARE OPEN. Can you prove it?

14. May 17, 2017

### Bashyboy

Suppose that $\{x\} \subset \Bbb{Q}$ is open. Since $x \in \{x\}$, there exists a $\epsilon > 0$ such that $(x-\epsilon,x + \epsilon) \subseteq \{x\}$, which cannot happen since the $\epsilon$-nbhd captures rationals outside of $\{x\}$.

Moreover, if the components, which we know are singletons, were indeed open, then the order topology on $\Bbb{Q}$ would be identical to the discrete topology, and therefore every set is clopen. This would mean, if I am not mistaken, that $S = \{x \in \Bbb{Q} ~|~ x^2 < 2\}$ is closed, which is clearly absurd since it has a limit point of $\sqrt{2}$.

15. May 17, 2017

### Dick

No, no, no. I mean can you prove that components are open in $\mathbb{R}^\omega$?