Suppose that
$B[0,1]$ := set of all bounded functions on $[0,1]$ equipped with the topology induced by the sup-norm
$C[0,1]$ :=set of all continuous functions on $[0,1]$.
Is $C[0,1]$ open in $B[0,1]$?
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Eric Stucky
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Topology
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Under what topology/metric? The answer is almost certainly no, in any case, but you need a topology or metric to define open. – PVAL-inactive Dec 28 '13 at 06:16
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Under uniform metric? No. But it is closed. – hot_queen Dec 28 '13 at 06:16
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Under sup metric..plz explain – Topology Dec 28 '13 at 06:18
3 Answers
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Hint: Let $f$ be a continuous function. Find a discontinuous function in any ball centered at $f$ (you only need to change the value of $f$ at one point). Why does this show $C[0,1]$ not open?
PVAL-inactive
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1@Mathematics: Perhaps you should explain which part is not clear, and what you tried to do in order to understand it. As it stands it's hard to know how to respond to your comment. (Note that an answer which offers a hint is not likely to be expanded to give you a complete answer that you could turn in with your homework, so if that's what you're waiting for, please go to some other site.) – Nate Eldredge Dec 28 '13 at 06:54
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Indeed you can show more.
As you probably know, a uniform limit of continuous functions is continuous. Once you have convinced yourself that convergence in the sup norm is the same as uniform convergence, you'll see this implies that $C[0,1]$ is closed in $B[0,1]$.
Now why does this imply that $C[0,1]$ is not open?
Verify that $B[0,1]$ (or indeed any topological vector space) is connected (indeed, path connected, and the paths can be very simple...)
In a connected topological space $X$, the only sets which are both open and closed are $\emptyset$ and $X$. Put another way, a proper nonempty subset which is closed can never be open.
Nate Eldredge
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If $C[0,1]$ is open in $B[0,1]$ then it is an open vector subspace of $B[0,1]$.
Lemma: $(V,||.||)$ be an NLS. $Y$ is open subspace of $V$ then $Y=V$.
Hence $C[0,1]$ cannot be open in $B[0,1]$
MathCosmo
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