Proving a Certain Set is Closed in the Uniform Topology

[Bashyboy]

Homework Statement

Let $Q = \{(x_1,x_2,...,) \in \mathbb{R}^\omega ~|~ \lim x_n = 0 \}$. I would like to show this set is closed in the uniform topology, which is generated by the metric $\rho(x,y) = \sup d(x_i,y_i)$, where $d$ is the standard bounded metric on $\mathbb{R}$.

Homework Equations

The Attempt at a Solution

Let $x = (x_n) \in \overline{Q} - Q$. Then $\lim x_n = \ell \neq 0$, and WLOG take $\ell > 0$. In my attempt, I mistakenly thought that $\prod (x_n - \ell, x_n + \ell)$ was a basis element for the product topology containing $x$, concluding from this that it must be open in the uniform topology. From there I proved that if $z = (z_n) \in \prod (x_n -\ell, x_n + \ell)$, then it cannot converge to $0$, which gives the desired contradiction.

My question is, is $\prod (x_n -\ell, x_n + \ell)$ in fact open in the uniform topology?

 

[haruspex]

Let $x = (x_n) \in \overline{Q} - Q$. Then $\lim x_n = \ell \neq 0$,

Is $\overline{Q}$ only convergent sequences?

 

[Bashyboy]

Yes. $\overline{Q}$ a subset of $\mathbb{R}^\omega$ which consists of all convergent real sequences.

 

[haruspex]

Yes. $\overline{Q}$ a subset of $\mathbb{R}^\omega$ which consists of all convergent real sequences.

But there is nothing in the problem statement that allows you to restrict attention that subset.

 

[Bashyboy]

But there is nothing in the problem statement that allows you to restrict attention that subset.

I don't quite follow...

I have a topological space $\mathbb{R}^\omega$, which I defined in a previous post; the space is endowed with the uniform topology, defined in the original post. Now, I have a subset $Q$ which I am trying to show is closed in this topology. Yes, you are right that I am restricting my attention to this particular subset, but I am doing so purposefully.

 

[haruspex]

I don't quite follow...

I have a topological space $\mathbb{R}^\omega$, which I defined in a previous post; the space is endowed with the uniform topology, defined in the original post. Now, I have a subset $Q$ which I am trying to show is closed in this topology. Yes, you are right that I am restricting my attention to this particular subset, but I am doing so purposefully.

My problem is that I so far see no connection between the thing to be proved and neither the $\overline Q$ subset nor the product topology you mention.
At the same time, I see no reference made to the metric nor to any usual definition of closure. If you are appealing to some standard results, please quote them.

I understand that (x_n-l) is a sequence, but I have no idea what you mean by $\prod (x_n - \ell, x_n + \ell)$. $(x_n - \ell, x_n + \ell)$ is a pair of numbers.$((x_n - \ell), (x_n + \ell))$ would be a pair of sequences. For a product topology we need pairs of sets.

Edit: I believe I have a fairly short proof from first principles.

 

[Bashyboy]

My problem is that I so far see no connection between the thing to be proved and neither the ¯¯¯¯QQ¯\overline Q subset nor the product topology you mention.
At the same time, I see no reference made to the metric nor to any usual definition of closure. If you are appealing to some standard results, please quote them.

I understand that (xn-l) is a sequence, but I have no idea what you mean by ∏(xn−ℓ,xn+ℓ)∏(xn−ℓ,xn+ℓ)\prod (x_n - \ell, x_n + \ell). (xn−ℓ,xn+ℓ)(xn−ℓ,xn+ℓ) (x_n - \ell, x_n + \ell) is a pair of numbers.((xn−ℓ),(xn+ℓ))((xn−ℓ),(xn+ℓ)) ((x_n - \ell), (x_n + \ell)) would be a pair of sequences. For a product topology we need pairs of sets.

Edit: I believe I have a fairly short proof from first principles.

Points in $\mathbb{R}^\omega$ can be regarded as points in a space, or they can be thought of as real convergent sequences in $\mathbb{R}$. When I write $(x_n)$ or $(x_n-\ell)$, I am referring to either one of these interpretations. When I write $\prod (x_n - \ell, x_n + \ell)$, on the other hand, I am writing a subset of the product space $\mathbb{R}^\omega$. This is in fact a basis element in the box topology. What I am trying to show is that it is open in the uniform topology, but I have been unsuccessful.

Every approach I have tried requires that I construct a basis element that containing no sequences that converge to zero. You say that you have a solution. Would you mind offering a hint?

 

[haruspex]

You say that you have a solution. Would you mind offering a hint?

The definition of closure in a metric space with which I am familiar is that the set contains all its limit points. So I took a sequence (of sequences) which, according to the given metric, converges, and showed ithat the sequence it converges to converges to zero.

 

[Bashyboy]

The definition of closure in a metric space with which I am familiar is that the set contains all its limit points. So I took a sequence (of sequences) which, according to the given metric, converges, and showed ithat the sequence it converges to converges to zero.

Unfortunately, I do not have that characterization of closure in a metric space. I only have the definition, which states that the closure of $A$ is the intersection of all closed sets containing $A$, and the general characterization that $x \in \overline{A}$ if and only if every open neighborhood of $x$ intersects $A$.

 

[haruspex]

Unfortunately, I do not have that characterization of closure in a metric space. I only have the definition, which states that the closure of $A$ is the intersection of all closed sets containing $A$,

Ok, but that only defines closure in terms of other closed sets, so cannot be fundamental.
See bullet 2 at
http://mathworld.wolfram.com/ClosedSet.html

and the general characterization that $x \in \overline{A}$ if and only if every open neighborhood of $x$ intersects $A$.

Ok, assuming the overline here means closure of, that matches the definition at bullet 4 of
http://mathworld.wolfram.com/SetClosure.html
In a metric space, if every open neighbourhood of x intersects set S then it follows that there exists an infinite sequence in S that converges to x. Thus x is a limit point of S. Since a closed set is one that is its own closure, the definition then says that a closed set contains all its limit points.