# Show set is open - technical difficulties

• quasar987
In summary, the problem asks to show that the closure of the interior of a closed convex set is equal to the set itself. The approach is to show that every point in the set is also in the closure of the interior, and every point in the closure of the interior is also in the set. This is proven by showing that the set of points in the interior of the set can be connected to any point in the set through a path that stays entirely within the set, demonstrating that the interior points are also in the closure. To show that every point in the closure of the interior is also in the set, a sequence of points along the axis of a "cone" shape formed by the set is constructed, and it is shown that
quasar987
Homework Helper
Gold Member
[SOLVED] Show set is open - technical difficulties

## Homework Statement

Let C be a closed convex set in a Banach space with non empty interior. Show that cl(int(C))=C (show that the closure of the interior of C is S itself).

## The Attempt at a Solution

Since int(C) is included in C, we have cl(int(C)) $\subset$ cl(C)=C.

For the other inclusion, let x be in C. I'm going to try to find a sequence in int(C) that converges to x. This will prove that C $\subset$ cl(int(C)).

Let y be in int(C) and B(y;r) be a ball centered on y and entirely contained in int(C). Since C is convex, the line d joining y and x is entirely contained in C. If I can show that d\{x} is entirely contained in int(C), then I will have won.

My idea is that not only is d contained in C, but so is the line from any point in B(y;r) to x! The set of all these lines make up a set the form of an ice cream cone where x is the tip and the ball B(y;r) is the ice cream ball.

So, I can either show that this ice cream cone minus {x} is open, or find explicitly for each point of d\{x} an open ball entirely contained in the ice cream cone.

In either case, how do we do that?!?

You can use a little of both. For each point in d/{x}, use the ice cream cone to show it has a ball around it in C, so is in int(C).

I really don't see how! What would be the first step?

How about considering s sequence of points along the axis of the cone, the line connecting y and x?

d/{x} is the set of points tx+(1-t)y with 0<=t<1. Label these points p(t). Let r be the radius of a ball B centered at y and contained in C. Then there is a ball B(t) of radius r(t) around p(t) that's contained in C. Because for any point z in B(t), consider the line passing through x and z. One convenient way to parameterize this is as:

$$q(s) = p(s) + \frac{1-s}{1-t} (z-p(t))$$

Then:

$$||y-q(0)|| = ||p(0)-q(0)|| = ||\frac{1}{1-t} (z-p(t))|| < \frac{r(t)}{1-t}$$

So we can pick r(t) so that q(0) lies in B, and so z must be in C.

Dick said:
How about considering s sequence of points along the axis of the cone, the line connecting y and x?

This is the idea, but we need to show first that the axis of the cone d\{x} is indeed in the interior of C.

StatusX said:
d/{x} is the set of points tx+(1-t)y with 0<=t<1. Label these points p(t). Let r be the radius of a ball B centered at y and contained in C. Then there is a ball B(t) of radius r(t) around p(t) that's contained in C. Because for any point z in B(t), consider the line passing through x and z. One convenient way to parameterize this is as:

$$q(s) = p(s) + \frac{1-s}{1-t} (z-p(t))$$

Then:

$$||y-q(0)|| = ||p(0)-q(0)|| = ||\frac{1}{1-t} (z-p(t))|| < \frac{r(t)}{1-t}$$

So we can pick r(t) so that q(0) lies in B, and so z must be in C.

I understand what you did but not quite how you did it. Specifically, how did you create that line q(s)? What were your guiding principles for coming up with this formula?

q(s) is just the line connecting x and z. Since z is close to p(t), this line should be "close" to p(s) (the line connecting x and p(t)), with the distance scaling with 1-s as you move farther away from x. The formula was written that way with this in mind so it'd be easy to find the actual point on q(s) that's close to p(0)=y.

## What does it mean when the show set is open but there are technical difficulties?

When the show set is open but there are technical difficulties, it means that the set is ready for the show to begin, but there are issues that need to be resolved before the show can start.

## What kind of technical difficulties can occur during a show set?

Technical difficulties can vary greatly and can include issues with lighting, sound, special effects, props, set pieces, or any other technical aspect of the show.

## How are technical difficulties handled during a show set?

The specific protocol for handling technical difficulties during a show set will vary depending on the production and the severity of the issue. In most cases, the stage manager or technical director will make the decision on how to proceed and communicate with the cast and crew.

## How long do technical difficulties typically last during a show set?

The duration of technical difficulties during a show set can vary greatly and is dependent on the nature of the issue. Some technical difficulties may be resolved quickly and the show can continue as planned, while others may require more time and may result in a delay or cancellation of the show.

## What can audience members do if there are technical difficulties during a show set?

If there are technical difficulties during a show set, audience members should remain calm and patient. In most cases, the issue will be resolved and the show will continue. If the show is cancelled or delayed, the theater staff will communicate this information to the audience and provide any necessary instructions or refunds.

• Calculus and Beyond Homework Help
Replies
3
Views
479
• Calculus and Beyond Homework Help
Replies
3
Views
668
• Calculus and Beyond Homework Help
Replies
1
Views
333
• Calculus and Beyond Homework Help
Replies
3
Views
643
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
2K
• Calculus and Beyond Homework Help
Replies
15
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
16
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
2K