Logical Point in Topological Problem

[Bashyboy]

Homework Statement

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.

Homework 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?

 

[Dick]

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?

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

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?

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.

 

[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.

 

[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.

 

[Dick]

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.

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.

 

[Bashyboy]

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

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.

 

[Dick]

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.

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).

 

[Bashyboy]

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

 

[Dick]

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

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?

 

[Bashyboy]

connected components are clopen.

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.

 

[Dick]

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.

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

 

[Bashyboy]

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

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.

 

[Dick]

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.

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

 

[Bashyboy]

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

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 $\sqrt{2}$.

 

[Dick]

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 $\sqrt{2}$.

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