
#1
Jun3011, 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)=supf(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 xy< s Now, Consider the open ball B(h,r/3). This is the set of all functions f such that fh < r/3 in the entire interval. I want to show that all the functions in this ball are discontinuous at the point y! 



#2
Jun3011, 09:13 AM

HW Helper
P: 3,353

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.




#3
Jun3011, 09:19 AM

Mentor
P: 16,624

Hi Oster!
Take xy<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] 



#4
Jun3011, 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 (ys,y+s)]
I'm not too comfortable with the convergence stuff. Sorry Gib Z. =( Thanks to both of you anyway! =D 



#5
Jun3011, 09:34 PM

HW Helper
P: 3,353

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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