why is (C[0,1],||.||4) not a complete space?
Since it is closed, is it because of the norm 4? I know it should be complete if for every sequence its limit is in the space.
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Closedness depends on the choice of topology. Your space is closed under topology induced by uniform norm ($\|f\| = \sup_{x\in[0,1]}|f(x)|$), but it is not closed under the topology induced by Lebesgue norm $$\|f\|_4 = \left(\int_0^1|f(x)|^4dx\right)^{1/4}.$$ In order to show that, you can find a counterexample to definition of closedness. For example, try to build a sequence of continuous functions $f_n$ such that they converge in the sense of the norm $\|\cdot\|_4$ to a discontinuous function $f$. You might want to start with a function $$f(x)=\begin{cases}1,&x\ge 0.5\\0,&x<0.5.\end{cases}$$
TZakrevskiy
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so under that topology the space would actually be open thus not complete? – user404056 May 30 '17 at 12:40
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1@SoHCahToha No, not necessarily open. Just "not closed". – TZakrevskiy May 30 '17 at 12:43
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is that sufficient to know that is it not complete since not closed hence not all limits of converging sequences will be in my set? – user404056 May 30 '17 at 12:46
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1@SoHCahToha exactly. – TZakrevskiy May 30 '17 at 12:49