Register to reply

Prove C[0,1] is closed in B[0,1] (sup norm)

by Oster
Tags: norm, prove
Share this thread:
Jun30-11, 09:04 AM
P: 85
So basically, my metric space X is the set of all bounded functions from [0,1] to the reals and the metric is defined as follows: d(f,g)=sup|f(x)-g(x)| where x belongs to [0,1].

I want to prove that the set of all discontinuous bounded functions, D[0,1] in X is open.

My attempt - Start with an arbitrary function h in D[0,1]. h is discontinuous.
=> There exists a point y in [0,1] and r>0 such that for all s>0
|f(x) - f(y)| > r when |x-y|< s

Now, Consider the open ball B(h,r/3). This is the set of all functions f such that |f-h| < r/3 in the entire interval. I want to show that all the functions in this ball are discontinuous at the point y!
Phys.Org News Partner Science news on
Scientists discover RNA modifications in some unexpected places
Scientists discover tropical tree microbiome in Panama
'Squid skin' metamaterials project yields vivid color display
Gib Z
Jun30-11, 09:13 AM
HW Helper
Gib Z's Avatar
P: 3,348
The easiest way here is actually to prove the statement in the title (that the set of continuous functions on [0,1] is closed). Use the fact that a set W is closed iff every convergent (wrt the relevant metric) sequence in W has limit in W.
Jun30-11, 09:19 AM
micromass's Avatar
P: 18,346
Hi Oster!

Take |x-y|<s. You know that |h(x)-h(y)|>r, and you'll need to find something for |f(x)-f(y)|

See if you can do something with this:


Jun30-11, 10:45 AM
P: 85
Prove C[0,1] is closed in B[0,1] (sup norm)

I tried micromass' method and got it! [|f(x)-f(y)|>r/3 in (y-s,y+s)]
I'm not too comfortable with the convergence stuff. Sorry Gib Z. =(
Thanks to both of you anyway! =D
Gib Z
Jun30-11, 09:34 PM
HW Helper
Gib Z's Avatar
P: 3,348
Well it can't hurt to see the other way, maybe it'll get you more used to it.

One of the most useful characterizations of a closed set is :

A set W is closed if and only if every convergent sequence in W has it's limit in W.

In R^n with the usual Euclidean metric, convergence you've seen before, it's the normal idea. In a general metric space (X,d) we say a sequence [itex] f_n [/itex] converges to [itex] f[/itex] if [itex] \displaystyle\lim_{n\to\infty} d(f_n, f) = 0 [/itex].

So in this case, we have to show that every convergent sequence of continuous functions on [0,1] has it's limit in C[0,1]. The norm we have here through is the supremum norm, so convergence in the supremum norm is just the concept of uniform convergence. It is a well known fact (and easy to prove on your own) from real analysis that if a sequence of continuous functions converges uniformly, then its limit is also a continuous function, which suffices to show C[0,1] is closed with respect to this supremum metric.

Register to reply

Related Discussions
Metric Space, closed ball is a closed set. prove this Calculus & Beyond Homework 5
Topology - Use Componentwise Convergence Criterion to prove closed ball closed. Calculus & Beyond Homework 3
Prove that the dual norm is in fact a norm Calculus & Beyond Homework 2
Prove that countable intersections of closed subset of R^d are closed Calculus & Beyond Homework 2