Show that f is a closed subset

[MaxManus]

Homework Statement

(X,d) is a metric space
B(x) is the set of all bounded functions on X

show that
f = {g $$\in$$ B(X) g is continious}
is a closed subset

The Attempt at a Solution

A hint of how to start?

 

[Mark44]

Homework Statement

(X,d) is a metric space
B(x) is the set of all bounded functions on X

show that
f = {g $$\in$$ B(X) g is continious}
is a closed subset

The Attempt at a Solution

A hint of how to start?

Definition of a closed subset?

 

[MaxManus]

Thanks
A subset of a metric space is closed if it contains all its boundary
points.

But I don't know how to apply it.

 

[Dick]

Thanks
A subset of a metric space is closed if it contains all its boundary
points.

But I don't know how to apply it.

Then you want to prove that if f_n is a sequence of bounded and continuous functions with a limit f, then f is bounded and continuous. What's the definition of f_n->f in your function space?

 

[MaxManus]

Thanks, but I don't understand what you are asking for.

If f_n has limit f then
abs(d(f_n, -f)) <= epsilon
for large enough n
Is this something I should use?

 

[Dick]

Thanks, but I don't understand what you are asking for.

If f_n has limit f then
abs(d(f_n, -f)) <= epsilon
for large enough n
Is this something I should use?

Definitely! First can you show f must be bounded?

 

[MaxManus]

show that d(f,a) <= M, where M <= infinity

d(f,a) <= d(f,f_n) + f(f_n,a) <= epsilon + M = M
Is this correct?

where "a" is in B(X)

 

[Dick]

show that d(f,a) <= M, where M <= infinity

d(f,a) <= d(f,f_n) + f(f_n,a) <= epsilon + M = M
Is this correct?

where "a" is in B(X)

Ok, now I realize I didn't ask enough questions. You say B(X) are bounded functions on X. Bounded functions from X to what? Are they functions from X->R where R is the real numbers? That's what I was assuming.

 

[MaxManus]

Yes, X-> R

 

[MaxManus]

More help?

 

[Dick]

Post 8 is sort of getting there, but I don't know what 'a' is doing in there. The notation could use a little work. d is the metric on x, so the distance between two points is d(x,y). The 'distance' between two functions f and g is max(|f(x)-g(x)|) for x in X. You can give that quantity a name if you want, but you probably don't want to call it 'd'. Bounded means |f(x)|<M for some constant M, right? Pick f_n 'close' to f. f_n is bounded, show f is bounded.

 

[MaxManus]

Thanks, I still don't understand.
I'm to show that:
|f(x)| < M
when:
|f_n| < M, for all n
|f_n -f| < epsilon, for n > N
But how?

 

[Dick]

Thanks, I still don't understand.
I'm to show that:
|f(x)| < M
when:
|f_n| < M, for all n
|f_n -f| < epsilon, for n > N
But how?

You don't have to show |f|<M. You have to show |f| is less than some constant. M+epsilon will do for that constant. And be a little careful, all you know is that each f_n is bounded by a constant. You don't know that they are all bounded by the same constant, until you happen to prove it.

 

[MaxManus]

|f_n -f| < epsilon
-epsilon<f_n - f < epsilon
-epsilon<M-f <epsilon
-M-epsilon <-f < epsilon - M
M + epsilon > f > M-epsilon
right?
But how does this show that f is closed?

 

[Dick]

|f_n -f| < epsilon
-epsilon<f_n - f < epsilon
-epsilon<M-f <epsilon
-M-epsilon <-f < epsilon - M
M + epsilon > f > M-epsilon
right?
But how does this show that f is closed?

Wrong. You can't just substitute M for f_n. f_n isn't equal to M. -M<f_n<M. And you aren't trying to show f is closed. You are trying to show f is bounded. If you get through this step then the next one is to show f is continuous. Then you can say your SET is closed.

 

[MaxManus]

But its right now now?:
|f_n -f| < epsilon
-epsilon<f_n - f < epsilon

-f_n-epsilon <-f < epsilon - f_
f_n + epsilon > f > f_n - epsilon
M +epsilon > f > -M-epsilon

 

[Dick]

But its right now now?:
|f_n -f| < epsilon
-epsilon<f_n - f < epsilon

-f_n-epsilon <-f < epsilon - f_
f_n + epsilon > f > f_n - epsilon
M +epsilon > f > -M-epsilon

The result is correct but the derivation isn't. You did the same thing you did last time. You substituted M for f_n. They AREN'T equal, are they??

 

[MaxManus]

But if
f_n + epsilon > f > f_n - epsilon
and:
-M<f_n<M
why not
M +epsilon > f > -M-epsilon
f_n can't be bigger than M or less than -M

 

[Dick]

But if
f_n + epsilon > f > f_n - epsilon
and:
-M<f_n<M
why not
M +epsilon > f > -M-epsilon
f_n can't be bigger than M or less than -M

Ok, I'll buy that now that you've explained your reasoning. So f is bounded. Now you want to show f is continuous.

 

[MaxManus]

Don't think I understand the problem;
f = {g $$\in$$ B(X) g is continuous}
Doesn't that mean that f is continuous?

 

[Dick]

Don't think I understand the problem;
f = {g $$\in$$ B(X) g is continuous}
Doesn't that mean that f is continuous?

Ok, there's a notation problem. We used f to be the function that's the limit of f_n. In the problem statement it looks like f was also used as the name of a set. Let's call the set F, so we don't confuse it with f as a function. They aren't the same.

 

[MaxManus]

If f is a function X -> Y, between two metric spaces.
f is continuous if for each epsilon > 0 there is a delta > 0 such that $$d_Y(f(x),f(y)) < \epsilon$$ for all x any y in X such that $$d_X(x,y) < \delta$$
But is this possible to use when you don't know the function?

 

[Dick]

If f is a function X -> Y, between two metric spaces.
f is continuous if for each epsilon > 0 there is a delta > 0 such that $$d_Y(f(x),f(y)) < \epsilon$$ for all x any y in X such that $$d_X(x,y) < \delta$$
But is this possible to use when you don't know the function?

Pick a point x. f_n is continuous, so you know there is a delta such that |f_n(x)-f_n(y)|<epsilon for all d(x,y)<delta. Now pick f_n to be close to f. Try to use that to say something about |f(x)-f(y)| for y where d(x,y)<delta.

 

[MaxManus]

|f_n(x)-f_n(y)|<epsilon for all d(x,y)<delta
|f_n -f| < epsilon2 for n > N
Triangle inequality
d(f(x),f(y)) <= d(f(x),f_n(x)) + d(f_n(x),f_n(y)) + d(f_n(y),f(y))< 2*epsilon2 + epsilon = epsilon3.

 

[Dick]

|f_n(x)-f_n(y)|<epsilon for all d(x,y)<delta
|f_n -f| < epsilon2 for n > N
Triangle inequality
d(f(x),f(y)) <= d(f(x),f_n(x)) + d(f_n(x),f_n(y)) + d(f_n(y),f(y))< 2*epsilon2 + epsilon = epsilon3.

Sure, that's basically it. But you don't want to pick a separate epsilon for |f_n-f|, just pick them both to be epsilon/3. Then you wind up with |f(x)-f(y)|<epsilon. And I'd use || for the metric on R, d is the metric on X.

 

[MaxManus]

But how do we go from here to knowing that F contains all its boundary points?

A subset of a metric space is closed if it contains all its boundary
points

 

[Dick]

But how do we go from here to knowing that F contains all its boundary points?

A subset of a metric space is closed if it contains all its boundary
points

Now you just have to think what all you've done means. You picked a sequence f_n in F that has a limit f. You want to show f is in F when the f_n are in F. You've used that the f_n are continuous and bounded to show f is continuous and bounded. Doesn't that mean f is in F?

 

[MaxManus]

You've used that the f_n are continuous and bounded to show f is continuous and bounded. Doesn't that mean f is in F?

I don't see it. But I know that if f is in F then it is closed

 

[Dick]

You've used that the f_n are continuous and bounded to show f is continuous and bounded. Doesn't that mean f is in F?

I don't see it. But I know that if f is in F then it is closed

Your definition of f being F isn't that f is 'closed', whatever that might mean. Functions aren't 'closed'. Sets are 'closed'. The definition of f being in F is that it's continuous and bounded!

 

[MaxManus]

I ment:
If {x_n} is a convergent sequence of elements in F, then the limit a =
lim x_n always belongs to F. = F is closed. I thought you used this rule, but no?