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Let $V$ be the vector space of the sequences whose values are contained in $\mathbb R$. Prove whether or not the following subsets $W$ $\subseteq$ $V$ are subspaces of $(V,+, \cdot)$:

a) $W=$ {($a_n$) $\in$ $V$: $\sum_{n=1}^\infty$ |$a_n$| < $\infty$

b) $W=$ {($a_n$) $\in$ $V$: $\lim_{n\to \infty}$ $a_n$ = $1$

c) $W=$ {($a_n$) $\in$ $V$: $\exists$ $\lim_{n\to \infty}$ $a_{2n}$

d) $W=$ {($a_n$) $\in$ $V$: # {$n: a_n \neq 0$} < $\infty$ }

I know that to prove a given subset is a subspace requires me to prove that it is closed under vector addition and multiplication, and that the zero vector belongs to the subset. However, I have not yet studied Calculus and am unfamiliar with sequences, so I am a little lost about how to go about proving this.

Any help is much appreciated!

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HINT

a) $W=$ {($a_n$) $\in$ $V$: $\sum_{n=1}^\infty$ |$a_n$| < $\infty$

  • let check directly that the three properties hold

b) $W=$ {($a_n$) $\in$ $V$: $\lim_{n\to \infty}$ $a_n$ = $1$

  • let consider $a_n=1$ and $b_n=1$ (or $a_n=0$)

c) $W=$ {($a_n$) $\in$ $V$: $\exists$ $\lim_{n\to \infty}$ $a_{2n}$

  • let check directly that the three properties hold

d) $W=$ {($a_n$) $\in$ $V$: # {$n: a_n \neq 0$} < $\infty$ }

  • let consider $a_n=0$
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  • $\begingroup$ Thank you gimusi! I appreciate your hints, I will try to apply them shortly and comment again. For now take a well-deserved +1. From what I'm gathering just at a glance, b) and d) aren't subspaces because it doesn't hold that the 0 vector belongs to W? $\endgroup$ – Jake S May 17 '18 at 20:09
  • $\begingroup$ @JakeS Yes exactly! The check on zero vector is always the first one to do! $\endgroup$ – gimusi May 17 '18 at 20:11
  • $\begingroup$ This question is quite old by now, but I was revisiting it as I study for a test, and I would like to know exactly what the proof for b) not being a subspace looks like. The others I'm quite convinced of, but I'm not sure of that one $\endgroup$ – Jake S May 25 '18 at 13:43
  • $\begingroup$ @JakeS simply note that $a_n=0\not\in W$ or that if $a_n,b_n\in W$ then $c_n=a_n+b_n \to 2$ then $c_n\not \in W$. $\endgroup$ – gimusi May 25 '18 at 13:49

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