Proving a Certain Set is Closed in the Uniform Topology

In summary, my attempt at a solution to the homework statement was to show that the set of all real convergent sequences in ##\mathbb{R}^\omega## is not closed in the uniform topology, but I incorrectly thought that the product of two such sequences was a basis element for this topology. After proving that if a given sequence z is in the product then it cannot converge to 0, I realized that the product topology is not open in the uniform topology.
  • #1
Bashyboy
1,421
5

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?
 
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  • #2
Bashyboy said:
Let ##x = (x_n) \in \overline{Q} - Q##. Then ##\lim x_n = \ell \neq 0##,
Is ##\overline{Q} ## only convergent sequences?
 
  • #3
Yes. ##\overline{Q}## a subset of ##\mathbb{R}^\omega## which consists of all convergent real sequences.
 
  • #4
Bashyboy said:
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.
 
  • #5
haruspex said:
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.
 
  • #6
Bashyboy said:
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 (xn-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.
 
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  • #7
haruspex said:
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?
 
  • #8
Bashyboy said:
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.
 
  • #9
haruspex said:
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##.
 
  • #10
Bashyboy said:
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

Bashyboy said:
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.
 

Related to Proving a Certain Set is Closed in the Uniform Topology

1. What is the definition of a closed set in the uniform topology?

In the uniform topology, a set is closed if it contains all of its limit points. This means that if a sequence of points from the set converges to a point outside of the set, that point must also be included in the set.

2. How can I prove that a set is closed in the uniform topology?

To prove that a set is closed in the uniform topology, you can use the definition of a closed set and show that the set contains all of its limit points. This can be done by assuming a point outside of the set and showing that it cannot be a limit point of the set.

3. What is the importance of proving a set is closed in the uniform topology?

Proving a set is closed in the uniform topology is important because it allows us to determine the behavior of the set under certain operations, such as limits and continuity. It also helps us to understand the structure of the space and how different sets interact with each other.

4. Can a set be both open and closed in the uniform topology?

Yes, a set can be both open and closed in the uniform topology. In fact, in a topological space, a set is open if and only if its complement is closed. This means that a set can be both open and closed if its complement is also open and closed.

5. How does the concept of closure relate to closed sets in the uniform topology?

The concept of closure is closely related to closed sets in the uniform topology. The closure of a set is the smallest closed set that contains it. Therefore, if a set is closed in the uniform topology, its closure is itself. However, if a set is not closed, its closure may be a larger set that includes all of its limit points.

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