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I am in the process of teaching some first year students about subspaces of $\mathbb{R}^{n}$ and need to stress that to check that $V\subseteq\mathbb{R}^{n}$ is a subspace, then you need to show that it is non empty. Specifically, they know that a subset $V\subseteq \mathbb{R}^{n}$ is a subspace if and only if the following three conditions hold:

  1. $V\neq\emptyset$;
  2. If $x,y\in V$, then $(x+y)\in V$; and
  3. If $x\in V$ and $\lambda\in\mathbb{R}$, then $\lambda\cdot x\in V$.

Is there actually an example of a subset $\emptyset=V\subseteq\mathbb{R}^{n}$ that is defined by a set of conditions such that if $x$ and $y$ were to satisfy those conditions, then $(x+y)$ and $\lambda\cdot x$ would also satisfy those conditions?

Edit: So specifically, has anyone got an example of a set of conditions on the coefficients of vectors in $\mathbb{R}^{n}$ such that

$V=\left\{\left(\begin{array}{c} a_{1} \\ a_{2} \\ \vdots \\ a_{n} \end{array}\right):a_{1},\ldots,a_{n}\text{ satisfy certain conditions}\right\}=\emptyset$

but that if $(a_{1},\ldots,a_{n})^{T}$ and $(b_{1},\ldots,b_{n})^{T}$ were to satisfy the conditions, then so would $a_{1}+b_{1}$ etc. So basically I want a set of conditions that are closed under addition and scalare multiplication, but together are inconsistent.

For example, $3\vert a_{1}$ is a condition that is closed under addition but not under scalar multiplication, $\vert a_{1}\vert \geq \vert a_{2}\vert$ is closed under scalar multiplication but not addition etc.

I hope that clarifies things a bit.

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If you chose the subset to be $\varnothing$, there are no elements to work with, at all. There is no reason to talk about "a subset", but rather "the" subset, since $\varnothing$ is unique. One usually sees $1.$ interchanged with ${\bf 0}\in V$, to stress the special role of the origin in vector spaces. –  Pedro Tamaroff Feb 27 '13 at 11:32
    
To have a subspace the span must form a group under vector addition. A group must contain an identity element, the zero vector $\mathbf 0$. A subspace spanned by ${\mathbf 0}$ is zero-dimensional (i.e. consists of one point). This is the smallest subspace that one can have. –  Tpofofn Feb 27 '13 at 11:41
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Oog, is there any way I can get you to reconsider that this is worth stressing? I'm not sure that many people find it so... at least, none of my teachers did... Algebraic objects and their subobjects simply always have nonempty underlying sets. –  rschwieb Feb 27 '13 at 14:49

5 Answers 5

up vote 2 down vote accepted

Well, if you want your cleverly written empty set to be closed under addition (seen simply by the given conditions, syntactically), then the $0$ will be very hard to exclude.

As Tara B noted, the emptyset, let be defined anyhow, is for sure closed under whatever operations. So, in that, any definition of the empty set is ok, and probably the easiest to present to students is

Let $V:=\emptyset$: this satisfies conditions 2. and 3. but not 1.

Another example, if you prefer (working over $\Bbb R$): $$V:=\left\{\pmatrix{x\\y} \in \Bbb R^2 : x=y,\ xy<0\right\} \ . $$

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That is exactly the kind of thing I'm looking for. The students' can easily see that it is closed under scalar multiplication. Showing that additivie closure does not hold is fairly easy, but by addidng a few more conditions, then it can be made a lot more complicated! Many thanks. –  David Ward Feb 27 '13 at 13:48
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@DavidWard: But additive closure does hold! –  Tara B Feb 27 '13 at 14:23
    
@TaraB My mistake (again) - sorry I'm having a bad day here! Thanks for all your helpful comments though! –  David Ward Feb 27 '13 at 15:49
    
@DavidWard: No problem! By the way, it always looks odd to me for an accepted answer to have zero votes (so I'll fix that now, because I think this answer is good). Surely if it's worth your tick, it's worth an upvote? –  Tara B Feb 27 '13 at 16:18
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I arrived to the question in time, when final edit was done. –  Berci Feb 27 '13 at 16:49

I think the problem lies in the fact the zero vector is always an element of a vector space.

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Sorry, I probably didn't make myself clear. I appreciate all the equivalent definitions etc. I've added an edit to clarify the situation. –  David Ward Feb 27 '13 at 11:37
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@DavidWard: I don't think this was just about giving an equivalent definition, it is the key reason why what you are asking for is not possible (see my answer). –  Tara B Feb 27 '13 at 12:01

As far as I understand, you are asking for a set of conditions $\{C_i\mid i\in I\}$ on vectors in $\mathbb{R}^n$, say, such that each $V_i := \{v\in \mathbb{R}^n\mid v \;\mathrm{satisfies} \;C_i\}$ is a vector space, but $\cap_{i\in I} V_i$, i.e. the set of all vectors in $\mathbb{R}^n$ satisfying all the conditions $C_i$, is empty.

This is not possible, because every vector space contains $0$, and hence the intersection of a collection of vector spaces always contains $0$ and is therefore non-empty.

EDIT: Or do you actually only want the $V_i$ to be closed under addition and scalar multiplication, rather than satisfy the full vector space axioms? In that case it is still not possible, since any set closed under addition and multiplication by scalars from a field still contains $0$.

FURTHER EDIT: What I think you possibly really mean to ask, though, is whether if $V=\emptyset$, then the conditions 2. and 3. are satisfied. And the answer to that is yes, trivially. The empty set has no elements, and so every statement of the form "For all $x\in \emptyset$, $P(x)$" is true.

YET ANOTHER EDIT/QUESTION: It just occurred to me that maybe what you are really asking for is a clever, non-obvious definition of the empty set as the set of vectors in $\mathbb{R}^n$ satisfying some set of conditions? The bit about closure under addition etc. is a distraction if that is the case, since, as I said, the empty set will always satisfy those anyway.

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Your final edit is precisely what I am looking for. Sorry if this wasn't clear. I think I may have found such a definition of the empty set. I just need to check it, and then I will post it to clarify things. –  David Ward Feb 27 '13 at 12:24
    
Ah, okay! Good to know. As I said, I was distracted from realising that was what you were getting at at first by other things in the question. I have to get on with other things now, but I'll be interested to look at your definition later. –  Tara B Feb 27 '13 at 12:34
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By the way, just to emphasise a point: you don't want the conditions to be 'individually closed under addition and scalar multiplication' as you say in your question, for the reasons I explained in my answer. –  Tara B Feb 27 '13 at 12:41

It is rather an algebra question than a linear algebra one. In fact a subset $H$ of a given group $G$ is defined as a subgroup if the identity element of $G$ is contained in $H$. Now a subspace is first of all a subgroup, so it must respect this request, which the empty set does not. On the other hand an empty set isn't even a group. The problem about an empty set is of a logic order. In fact you cannot affirm for example:$$\exists e \in \emptyset : \forall x \in \emptyset, x\star e=e\star x=x,$$ because there does not exist an $e\in\emptyset.$ But on the other hand you can't affirm the converse, because you don't have $x\in\emptyset$. So you can affirm everything of an empty set, which is much alike affirming nothing.

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You can try the intuitive approach that a vector subspace may be regarded as the space of solutions of a system of homoegeneous equations, which then can never be empty (you always have the trivial solution $\vec{v}=0$). Note that a solution of non-homogeneous system is not a vector space but an affine space.

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