Why C([0,1]) and C_{o} are not isomorphic?

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Can anyone explain why C([0,1]) and C_{o} [tex]\oplus C [/tex][tex]\oplus C [/tex] (where C=complex ) are not isomorphic?
 

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  • #2
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Your question is not clear to me. Not isomorphic as what? Which category? C([0,1]) looks like the space of continuous functions on the unit interval. With what structure? Banach space? What is C_{o}?
 
  • #3
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Sorry for not being clear.

Yes, C([0,1]) is as a Banach space and C_{o} is the space of function vanishing at infinity.
 
  • #4
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I am trying to think, but I got confused. We are on on a compact interval and C_{o} does not make much sense for me in this situation. Say, I take f(x)=|x-1/2|. It is continuous, vanishes at 1/2 and so it is in C_{o} (can be made arbitrarily small outside a compact set)!
 
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  • #5
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Sorry, I forgot to put in (0,1). For some reason, I think that I did. (Texing is not going very well for me.) So, It's the space of function on (0,1) vanishing at infinity.
 
  • #6
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But for continuous functions on an open interval which norm are you taking?
 
  • #7
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I assume that we are taking the sup norm. Is there a problem?
 
  • #8
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Yes, there is a problem. For instance 1/x is a continuous function on the open interval (0,1).
 
  • #9
Office_Shredder
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The only definitions I've seen about functions vanishing at infinity require you to have points that go to infinity. How are you defining it on the interval (0,1)?
 
  • #10
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The only definitions I've seen about functions vanishing at infinity require you to have points that go to infinity. How are you defining it on the interval (0,1)?

I guess like in http://en.wikipedia.org/wiki/Continuous_functions_on_a_compact_Hausdorff_space" [Broken] in "Generalizations".

I guess he will replace now C(X) by "continuous bounded functions", but then we have sin(1/x)
which is continuous bounded, but has no limit at x=0 - which will spoil the isomorphism. My guess is that these two complex C were supposed to be the limits at both ends.
 
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  • #11
Office_Shredder
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From a first glance, it seems that vanishing at infinity on (0,1) just means continuous functions on [0,1] which are zero at 0 and 1. Just looking at intervals of the form (1/n, 1-1/n), for each epsilon f(x) can only be larger than epsilon on finitely many of these intervals, which means that looking at the limit at x goes to zero or 1 must be zero. Is that wrong?
 
  • #12
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From a first glance, it seems that vanishing at infinity on (0,1) just means continuous functions on [0,1] which are zero at 0 and 1. Just looking at intervals of the form (1/n, 1-1/n), for each epsilon f(x) can only be larger than epsilon on finitely many of these intervals, which means that looking at the limit at x goes to zero or 1 must be zero. Is that wrong?

I think this is the case. The mistake of the OP was that he thought that every bounded continuous function has finite limits at both end. You would subtract then the straight line connecting these segments and you get vanishing at the ends.
 
  • #13
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I was trying to figure out why the sequence:

0 --> C_{o}((0,1)) --> C[([0,1]) --> C[tex]\oplus[/tex]C --> 0

is exact but does not split. I got confused thinking that if the middle is not isomorphic to the direct sum of two sides then the sequence does not split. This is only the case when the one on the left is unital.
 

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