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

by Oster
Tags: norm, prove
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Oster
#1
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!
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Gib Z
#2
Jun30-11, 09:13 AM
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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.
micromass
#3
Jun30-11, 09:19 AM
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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:

[tex]r<|h(x)-h(y)|=|(h(x)-f(x))+(f(x)-f(y))+(f(y)-h(y))|[/tex]

Oster
#4
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
#5
Jun30-11, 09:34 PM
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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.


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