Proving a set in 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.

Homework Equations

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.

 

[micromass]

Hi l888l888l888!

Let's prove that D_n is open. Pick an f in D_n. 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

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

 

[l888l888l888]

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?

 

[micromass]

i think you meant prove Dn is closed. but anyway...

Yes, I meant to prove that the complement of D_n 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?

 

[l888l888l888]

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)|?

 

[micromass]

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

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

 

[l888l888l888]

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?

 

[micromass]

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?

 

[l888l888l888]

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?

 

[micromass]

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...

 

[l888l888l888]

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

 

[micromass]

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.

 

[l888l888l888]

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.

 

[micromass]

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.