- #1
elias001
- 13
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I have 3 questions concerning trying to prove open and closed sets for specific sequence spaces, they are all kind of similar and somewhat related. I thought i would put them all in one thread instead of having 3 threads.
1) Given y=(y[itex]_{n}[/itex]) [itex]\in[/itex] H[itex]^{∞}[/itex], N [itex]\in[/itex]N and ε>0, show that the set A={x=(x[itex]_{n}[/itex])[itex]\in[/itex] H[itex]^{∞}[/itex]:lx[itex]_{k}[/itex]- y[itex]_{k}[/itex]l<ε, for k=1,2,...N} is open in H[itex]^{∞}[/itex]
2) Show that c[itex]_{0}[/itex] is a closed subset of l[itex]_{∞}[/itex] [Hint: if (x[itex]^{(n)}[/itex]) is a sequence (of sequences!) in c[itex]_{0}[/itex] converging to x [itex]\in[/itex] l[itex]_{∞}[/itex], note that lx[itex]_{k}[/itex]l [itex]\leq[/itex] lx[itex]_{k}[/itex] - x[itex]^{n}_{k}[/itex]l + lx[itex]^{n}_{k}[/itex]l and now choose n so that lx[itex]_{k}[/itex] - x[itex]^{n}_{k}[/itex]l is small independent of k.]
3) show that the set B={x [itex]\in[/itex] l[itex]_{2}[/itex]: lx[itex]_{n}[/itex]l[itex]\leq[/itex]1/n, n=1,2,..} is not an open set. [hint: is the ball B(0,r) a subset of B.]
for question 1
The metric for H^infinity is A,
d(x,y)=Ʃ[itex]^{∞}_{i=1}[/itex]2[itex]^{-n}[/itex]lx[itex]_{n}[/itex]-y[itex]_{n}[/itex]l
so if m is in A, then I can enclosed a ball B(m,[itex]\delta[/itex]) of radius delta around m where [itex]\delta[/itex]=M[itex]_{k}[/itex]+1/2^k where M[itex]_{k}[/itex]=max{lx[itex]_{k}[/itex]-y[itex]_{k}[/itex]l, k=1,..,N} would that work?
For question 2,
in the hint, where it says to choose n, so that... how do i pick n so that lx[itex]_{k}[/itex] - x[itex]^{n}_{k}[/itex]l is small.
I know that lx[itex]_{k}[/itex]l≤ lxl, but you have a sequence x[itex]^{n}_{k}[/itex] converging to x, is d(x[itex]^{n}_{k}[/itex], x) ≥ d(x[itex]^{n}_{k}[/itex], x[itex]_{k}[/itex])?? If so, how can i use this fact to pick n?
for question 3.
From the hint where it asks whether the open ball of radius r around 0 is a subset of B. In this case, would i have to specific how large r would be so that the radius of B(0,r) would contains elements that are not in B. If so, how do i specific r. The minkowski inequality is not of any help.
1) Given y=(y[itex]_{n}[/itex]) [itex]\in[/itex] H[itex]^{∞}[/itex], N [itex]\in[/itex]N and ε>0, show that the set A={x=(x[itex]_{n}[/itex])[itex]\in[/itex] H[itex]^{∞}[/itex]:lx[itex]_{k}[/itex]- y[itex]_{k}[/itex]l<ε, for k=1,2,...N} is open in H[itex]^{∞}[/itex]
2) Show that c[itex]_{0}[/itex] is a closed subset of l[itex]_{∞}[/itex] [Hint: if (x[itex]^{(n)}[/itex]) is a sequence (of sequences!) in c[itex]_{0}[/itex] converging to x [itex]\in[/itex] l[itex]_{∞}[/itex], note that lx[itex]_{k}[/itex]l [itex]\leq[/itex] lx[itex]_{k}[/itex] - x[itex]^{n}_{k}[/itex]l + lx[itex]^{n}_{k}[/itex]l and now choose n so that lx[itex]_{k}[/itex] - x[itex]^{n}_{k}[/itex]l is small independent of k.]
3) show that the set B={x [itex]\in[/itex] l[itex]_{2}[/itex]: lx[itex]_{n}[/itex]l[itex]\leq[/itex]1/n, n=1,2,..} is not an open set. [hint: is the ball B(0,r) a subset of B.]
for question 1
The metric for H^infinity is A,
d(x,y)=Ʃ[itex]^{∞}_{i=1}[/itex]2[itex]^{-n}[/itex]lx[itex]_{n}[/itex]-y[itex]_{n}[/itex]l
so if m is in A, then I can enclosed a ball B(m,[itex]\delta[/itex]) of radius delta around m where [itex]\delta[/itex]=M[itex]_{k}[/itex]+1/2^k where M[itex]_{k}[/itex]=max{lx[itex]_{k}[/itex]-y[itex]_{k}[/itex]l, k=1,..,N} would that work?
For question 2,
in the hint, where it says to choose n, so that... how do i pick n so that lx[itex]_{k}[/itex] - x[itex]^{n}_{k}[/itex]l is small.
I know that lx[itex]_{k}[/itex]l≤ lxl, but you have a sequence x[itex]^{n}_{k}[/itex] converging to x, is d(x[itex]^{n}_{k}[/itex], x) ≥ d(x[itex]^{n}_{k}[/itex], x[itex]_{k}[/itex])?? If so, how can i use this fact to pick n?
for question 3.
From the hint where it asks whether the open ball of radius r around 0 is a subset of B. In this case, would i have to specific how large r would be so that the radius of B(0,r) would contains elements that are not in B. If so, how do i specific r. The minkowski inequality is not of any help.