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?

 

[Dick]

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?

That's what you just did. Translate the statement '(blank) is an element of F' to '(blank) is continuous and bounded' where (blank) is anything. Because that's the definition of F!

 

[MaxManus]

Thanks for your patience. It took some days, but I get that the two are the same thing now.