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

• Oster
In summary, we want to prove that the set of all discontinuous bounded functions, D[0,1] in X is open. To do this, we consider an arbitrary function h in D[0,1] and show that there exists an open ball around h such that all functions within the ball are also discontinuous at a certain point y in [0,1]. Another approach to proving D[0,1] is open is to show that the set of continuous functions on [0,1] is closed, using the characterization that a set is closed if every convergent sequence in the set has its limit in the set. This can be shown by using the concept of uniform convergence, which states that a sequence of continuous
Oster
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!

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.

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:

$$r<|h(x)-h(y)|=|(h(x)-f(x))+(f(x)-f(y))+(f(y)-h(y))|$$

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

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 $f_n$ converges to $f$ if $\displaystyle\lim_{n\to\infty} d(f_n, f) = 0$.

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.

## 1. What is the definition of a closed set?

A closed set is a subset of a metric space that contains all of its limit points. In other words, for any sequence of points within the set that converges to a point outside of the set, the point is still considered to be within the set.

## 2. How do you prove that a set is closed?

To prove that a set is closed, you must show that it contains all of its limit points. This can be done by showing that for any sequence of points within the set that converges to a point outside of the set, the point is still contained within the set. Another way to prove a set is closed is by showing that its complement (the set of points not in the set) is open.

## 3. What is the definition of the sup norm?

The sup norm, also known as the supremum norm or the uniform norm, is a type of metric used to measure the distance between two functions. It is defined as the maximum absolute value of the difference between the two functions over a given interval.

## 4. How do you prove that C[0,1] is a subset of B[0,1]?

To prove that C[0,1] is a subset of B[0,1], you must show that every function in C[0,1] is also in B[0,1]. This can be done by showing that the sup norm of the function is finite, meaning that the function does not grow infinitely large over the interval [0,1].

## 5. Can you provide an example of a function in C[0,1] that is not in B[0,1]?

Yes, an example of a function in C[0,1] that is not in B[0,1] is f(x) = 1/x on the interval [0,1]. This function is continuous on the interval and therefore in C[0,1], but its sup norm is infinite, making it not in B[0,1].

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