# Proving a set in closed and nowhere dense

• l888l888l888
In summary: Right, I meant that there exists a function g such that the difference between f and g is less than some epsilon, and the slope between two close points t and t+h is greater than n. This is the definition of a function being close to another function, with the added requirement of a large slope.In summary, we need to prove that for every function f, there exists a function g that is close to f and has a large slope between two close points, in order to show that Dn is closed and nowhere dense.
l888l888l888

## Homework Statement

Dn={f in C([0,1]) : there exists t in [0,1] for every h in R/{0}, abs((f(t+h)-f(t))/h) <=n}
prove the Dn's are closed nowhere dense sets. A subset of some set A is closed in A if its complement is open in A.

## The Attempt at a Solution

i first defined the complement of Dn as the set in which for all t in [0,1], abs((f(t+h)-f(t))/h)>n. Is trying to prove closed first and then nowhere dense.

Hi l888l888l888!

Let's prove that Dn is open. Pick an f in Dn. You know that

$$|f(t+h)-f(t)|>n|h|$$

You'll need to find an epsilon such that for all g with $\|f-g\|_\infty$ holds that

$$|g(t+h)-g(t)|>n|h|$$

Start by working out

$$|f(t+h)-f(t)|=|(f(t+h)-g(t+h))+(g(t+h)-g(t))+(g(t)-f(t))|$$

i think you meant prove Dn is closed. but anyway...
f(t+h)−f(t)|=|(f(t+h)−g(t+h))+(g(t+h)−g(t))+(g(t)−f(t))|
<=|f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)|. is that what you meant?

l888l888l888 said:
i think you meant prove Dn is closed. but anyway...

Yes, I meant to prove that the complement of Dn is open.

f(t+h)−f(t)|=|(f(t+h)−g(t+h))+(g(t+h)−g(t))+(g(t)−f(t))|
<=|f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)|. is that what you meant?

Yes, and you know that

$$n|h|<|f(t+h)-f(t)|$$

and $\|f-g\|_\infty<\epsilon$. So what can you deduce from this?

well if n|h|<|f(t+h)−f(t)| and if |f(t+h)−f(t)|<=|f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)| then n|h|< |f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)|. and if sup |f-g|< epsilon then n|h|< |f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)|< 2epsilon + |g(t+h)−g(t)|?

l888l888l888 said:
well if n|h|<|f(t+h)−f(t)| and if |f(t+h)−f(t)|<=|f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)| then n|h|< |f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)|. and if sup |f-g|< epsilon then n|h|< |f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)|< 2epsilon + |g(t+h)−g(t)|?

Indeed, but let's refine this result. So you know that

$$n|h|<|f(t)+f(t+h)|$$

So there is actually a number a such that

$$n|h|+a\leq|f(t)+f(t+h)|$$

So, by the same reasoning, you have

$$n|h|+a<2\epsilon+|g(t+h)-g(t)|$$

So, can you find an epsilon small enough such that

$$n|h|<|g(t+h)-g(t)|$$

Right! to be more specific are we letting epsilon to be a/2? also for the nowhere dense part. there are many theorems to use in my book about proving something is nowhere dense. for instance we have " a subset is nowhere dense if int(cl(H)) = empty set or iff T- Cl(H) is dense. which shall i use or do u suggest to use another theorem?

l888l888l888 said:
Right! to be more specific are we letting epsilon to be a/2?

Indeed.

also for the nowhere dense part. there are many theorems to use in my book about proving something is nowhere dense. for instance we have " a subset is nowhere dense if int(cl(H)) = empty set or iff T- Cl(H) is dense. which shall i use or do u suggest to use another theorem?

Since H is closed, you'll only need to prove that int(H) is closed or that T/H is dense.
So, what do we need to prove?? We need to show that very every function f, there is a function g such that $\|f-g\|_\infty$ and such that

$$\frac{|(g(t+h)-g(t)|}{|h|}\geq n$$

Can you see graphically what must happen? We need to find a function g very close to f such that the slope between two close points t and t+h is big.

Let's do this for a simple case first: let's take f=0. Then we need to find a function g which is always close to zero, but the slope between two very close points must become large. Can you find me such a function?

so what your saying is basically we need g=mx+b where m the slope is sufficiently large but g~0 (g is close to 0). so mx+b~0...? graphically wen i draw it out its hard to explain it on here. g has to intercept the y-axis at a negative number?

l888l888l888 said:
so what your saying is basically we need g=mx+b where m the slope is sufficiently large but g~0 (g is close to 0). so mx+b~0...? graphically wen i draw it out its hard to explain it on here. g has to intercept the y-axis at a negative number?

Hmm, why do you say that we need g=mx+b?? All we need is that for every point t, there is a point t+h, such that if we draw the line [g(t),g(t+h)] then it's slope is sufficiently large.

A way to accomplish this is to let g occilate fast...

AH! Ok so considering a function g that oscillates about the x-axis (y=0)? now what?

l888l888l888 said:
AH! Ok so considering a function g that oscillates about the x-axis (y=0)? now what?

Now try to do this for a general f.

so for every function f there is a functions g that oscillates about f and the slope between g(t+h) and g(t) is sufficiently large and g is still close to f. Also, when you said,
"We need to show that very every function f, there is a function g such that ∥f−g∥∞ and such that

|(g(t+h)−g(t)|/|h|≥n" did you mean there is a function g st ||f-g||<some epsilon. that would be the closeness of f and g you are referring to.

l888l888l888 said:
so for every function f there is a functions g that oscillates about f and the slope between g(t+h) and g(t) is sufficiently large and g is still close to f. Also, when you said,
"We need to show that very every function f, there is a function g such that ∥f−g∥∞ and such that

|(g(t+h)−g(t)|/|h|≥n" did you mean there is a function g st ||f-g||<some epsilon. that would be the closeness of f and g you are referring to.

Yes, for every epsilon, there is a function g such that $\|f-g\|_\infty$ and such that

$$\frac{|g(t+h)-g(t)|}{|h|}\geq n$$.

And the function g which we will take is a function that oscillates fast about f.

## 1. What does it mean for a set to be closed and nowhere dense?

A set is considered closed if it contains all of its limit points. It is nowhere dense if its closure does not contain any non-empty open subset. In other words, the set is not dense in any part of the space.

## 2. How do you prove that a set is closed and nowhere dense?

To prove that a set is closed and nowhere dense, you can show that its closure is equal to the set itself and that it does not contain any non-empty open subset. In other words, you can show that the set is not dense in any part of the space.

## 3. What is the importance of proving a set is closed and nowhere dense?

Proving that a set is closed and nowhere dense is important in many areas of mathematics, including analysis, topology, and geometry. It helps to characterize the properties of a set and understand its behavior in a larger space.

## 4. Can a set be closed but not nowhere dense or vice versa?

Yes, a set can be either closed or nowhere dense, but not both at the same time. For example, a closed interval in the real line is not nowhere dense, while the rational numbers in the real line are nowhere dense but not closed.

## 5. What are some examples of sets that are closed and nowhere dense?

Some examples of sets that are closed and nowhere dense include finite sets of real numbers, the Cantor set, and the set of irrational numbers in the real line. These sets have no interior points and are not dense in any part of the space.

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