# Function Spaces

1. Apr 5, 2005

### Ed Quanta

Let T be an element of B[0,1] be the set V=[B[0,1];f(1)=2]. Prove that T is closed (in metric space B[0,1]).

I am not sure if it is obvious since I am new to this stuff but B[0,1] is an open ball I believe.

My question is how do I find the complement of V. If I could define B/T then I am hoping it will follow easily from the definitions of open set that this is an open set and my proof will be complete. Is B/T=[B[0,1]; f(1)>2 V f(1)<2]? Very confused.

2. Apr 5, 2005

### PBRMEASAP

What is B[0,1], the set of functions on [0,1] into R? Are they continuous? What is the metric?

3. Apr 5, 2005

### mathwonk

standard stuff would be something like this:

1) B[0,1] is the set of continuous functions from [0,1] to the reals

or maybe because there is a B there, it is just bounded functions.

2) the topology on B[0,1] is given by say the sup norm, i.e. the distance from f to g is the furthest apart any two values f(t), g(t) ever get for t in [0,1].
3) then prove the evaluation map taking f to f(1) is continuous.

4) then the inverse image of 2 under the evaluation map is clsoed since 2 is closed in R.

4. Apr 5, 2005

### Ed Quanta

Thanks a lot mathwonk

5. Apr 9, 2005

### Ed Quanta

What if I were to try to solve this by taking the complement of V and using open or closed balls? Is there any way to prove it this way? B[0,1] is the set of bounded functions by the way.