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Consider a subspace $H$, where $H= \{(x_1, x_2, \dots)\in \mathbb{F} \ | \ x_1 = 0 \} $ is a subspace of the "sequence room" $\mathbb{F}$ (the room $\mathbb{F}$ defined to contain all real sequences of the type $\mathbf{x}= \{x_j\}^\infty_{j=x}=(x_1,x_2,x_3,\dots )$ ).

In my notes I have as an example that, if $x_1 = 2$, then $H$ is not a subspace of $\mathbb{F}$. Why is that?

Is it simply because, with $x_1$ defined as nonzero, then the zero vector can not be in $H$? Hence $(0,0,0,\dots) \notin H$, and thus one of the criteria for a subspace is not satisfied?

Is that it, or is there something else I have missed?

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In fact, if you require $x_1=2$, then none of the criteria are satisfied, except non-emptiness: the sum of two sequences with first term $2$ is a sequence whose first term is not $2$, so the sum of two vectors in that set is not in the set. Multiplying a sequence with first term 2 by any scalar other than 1 will result in a sequence whose first term is not 2, so the set is also not closed under scalar multiplication. The zero vector being in the set is necessary for being a subspace, but not sufficient and not always part of the "defining criteria". –  Arturo Magidin Aug 29 '11 at 16:14
    
@Arturo Thank you! –  Jodles Aug 29 '11 at 16:41
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2 Answers

up vote 5 down vote accepted

You're absolutely correct: the set $$H=\{(x_1,x_2,\ldots)\in F\mid x_1=2\}$$ doesn't contain the zero element $(0,0,\ldots)$, which means $H$ is not a subspace. Other reasons why $H$ is not a subspace is that it is not closed under vector addition, for example: $$(2,0,0,\ldots)+(2,0,0,\ldots)=(4,0,0,\ldots)\notin H$$ or scalar multiplication, for example: $$3(2,0,0,\ldots)=(6,0,0,\ldots)\notin H$$

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What you suspect is exactly correct! In order to be a subspace, it must contain the identity element of addition, which is in this case the zero vector $(0,0,0...)$. If $x_1=2$ then the zero vector can not possible be in the "subspace", so it's not really a subspace after all.

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