Totally Bounded in a Function Space

In summary, the conversation discusses the concept of total boundedness in relation to metric spaces and the set of discontinuities of a function. The main question is whether showing D(f) is totally bounded is enough or if there is something else that needs to be shown. One suggestion is to show that if a point is continuous for all functions fn, then it is also continuous for f. This would prove the inclusion of the set of discontinuities of f in the union of the sets of discontinuities of the functions fn.
  • #1
jdcasey9
28
0

Homework Statement



Let (X,d), (Y,p) be metric spaces, and let f,fn: X -> Y with fn->f uniformly on X. Show that D(f)c the union of D(fn) from n=1 to n=infinity, where D(f) is the set of discontinuities of f.

Homework Equations





The Attempt at a Solution



Ok, so this looks pretty close to just a straight forward problem of total boundedness, but I'm assuming there is some difference that I'm overlooking. If we just try to show that D(f) is totally bounded we get:

Let E>0. Take B(E, f1), ... , B(E, fn) (where E is the radius and f is the center) over the interval f = (f1,fn) where fi+1 - fi = E/2. Therefore, we have an E-net covering D(f) and D(f) is totally bounded.

Is there something more to show/do or am I way off?
 
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  • #2
Ummm, what does total boundedness has to do with this?

Cant you just show:
If x is a point such that all f_n is continuous in x, then f is continuous in x.

That would prove that

[tex]\bigcap_n{X\setminus D_{f_n}}\subseteq X\setminus D_f [/tex]
 
  • #3
Really? Are you sure? What we are showing matches my definition of totally bounded nearly verbatim.
 

FAQ: Totally Bounded in a Function Space

What does it mean for a function space to be totally bounded?

A function space is considered totally bounded if every sequence in the space has a Cauchy subsequence. This means that the space is "compact" in the sense that every sequence in the space has a convergent subsequence.

How is the concept of total boundedness related to the completeness of a function space?

A function space is complete if every Cauchy sequence in the space converges to a point in the space. Total boundedness is a necessary condition for completeness, but it is not sufficient. This means that a totally bounded space may or may not be complete, but a complete space must be totally bounded.

Can a function space be totally bounded but not compact?

Yes, a function space can be totally bounded but not compact. This means that while every sequence in the space has a convergent subsequence, the space itself may not be "small" enough to be considered compact. In other words, it may have an infinite number of elements.

How does the concept of total boundedness apply to metric spaces?

The concept of total boundedness can also be applied to metric spaces, where it is equivalent to the notion of "boundedness". In a metric space, a set is totally bounded if it can be covered by finitely many balls of a given radius. This is similar to the definition in a function space, where a sequence can be "covered" by a Cauchy subsequence.

Can a function space be totally bounded in one norm but not in another?

Yes, it is possible for a function space to be totally bounded in one norm but not in another. This is because different norms can induce different topologies on the same space, and therefore, the concept of total boundedness may differ. For example, a function space may be totally bounded in the L^1 norm but not in the L^2 norm.

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