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Define $\|\cdot\|\colon V \rightarrow \mathbb{R} $ by $\|a\| = \sum 2^n |a_n| $, where $V$ denotes the vector space of all sequences $a=(a_1,a_2,a_3,\dotsc)$ of real numbers such that $\sum 2^n |a_n| $ converges.

Which of the following are true?

1) $V$ contains only the sequence $(0,0,\dotsc)$

2) $V$ is finite dimensional

3) $V$ has a countable linear basis

4) $V$ is a complete normed space

My answer: it is already given that $\sum 2^n |a_n| $ converges. That means $V$ contain contains only the sequence $(0,0,\dotsc)$ so option 1) is correct.

Option 2 is incorrect because $\mathbb{R}$ is uncountable, as continuous image of uncountable is uncountable.

I'm confused about option 3 and option 4.

Please help me.

Any hints/solution will be appreciated.

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  • 2
    $\begingroup$ Hints: 1, 2 and 3 are incorrect. $\endgroup$
    – egreg
    Commented Jul 15, 2018 at 20:59
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    $\begingroup$ Why is 1) correct? It means $dim V=0$ $\endgroup$
    – Naj Kamp
    Commented Jul 15, 2018 at 21:00
  • $\begingroup$ how ?? im not getting @egreg $\endgroup$
    – jasmine
    Commented Jul 15, 2018 at 21:00

2 Answers 2

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Option 1 is incorrect, because all sequences having only finitely many nonzero entries belong to $V$. This also easily shows that 2 is incorrect.

Also option 3 is incorrect, because a complete normed vector space cannot have a countably infinite basis (either finite or uncountable).

On the other hand, option 4 is correct, because…

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  • $\begingroup$ @ egreg ..can u little more elaborate,,,why not infinitely many non zero entries belong to V?? $\endgroup$
    – jasmine
    Commented Jul 15, 2018 at 21:07
  • $\begingroup$ Messi, can you show us some non trivial elements of $V$? $\endgroup$
    – Naj Kamp
    Commented Jul 15, 2018 at 21:21
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    $\begingroup$ $(1,0,0,\cdots)$ is a non-zero element of $V.$ As is $(1,a,a^2,a^3,\cdots)$ when $|a|<1/2$ $\endgroup$ Commented Jul 15, 2018 at 21:49
  • $\begingroup$ @egreg can space have Shauder basis? $\endgroup$
    – user464147
    Commented Jul 16, 2018 at 16:30
  • $\begingroup$ @FailedtobeaMathematician This space definitely has one. $\endgroup$
    – egreg
    Commented Jul 16, 2018 at 17:11
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Hint:

The map $\ell^1 \to V$ given by $$(x_n)_n \mapsto \left(\frac{x_n}{2^n}\right)_n$$ is an isometric isomorphism.

Therefore just check what properties hold for $\ell^1$ (only option $4$).

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