While reading text, I had a question which I can not resolve by myself. Please Help me! it reads, The empty set( a vector space with no elements) is denoted as & (This symbol doesn't matter for the sake of argument, I don't know how to write the Zero with a line in the middle). I can understand what they mean by empty set. It must be somewhat like this; Set V = { }. But can such set with no elements has all the quality of being a vector space? If so, how can one show it does indeed meets all the ten axioms of vector space?
It seems to me that a vector space ought to contain vectors, so the empty set probably doesn't qualify.
The thing with maths is that the empty set can be a FOO since there are no elements of the empty set for whcih the definitions of FOO are false. BUt, that said, I don't think this case is one of them, there is for instance no additive identity. Which text is this?
Hm Since the empty set is a subset of every set, I think one can still say it's a vector space. The statement [tex]x\in\emptyset[/tex] is always wrong, so you can derive any statment you wish, in this case the axioms for vector spaces, which will apply to the empty vector space. (In fact, everything can follow from a wrong statement)
Vector spaces need a zero vector (an additive identity) just like groups need an identity element. So empty sets cannot be vector spaces.
Of course, almost every time I define these things for students I declare the underlying set to be not empty and that removes any doubt.
OK, I looked it up. The definition presupposes that a subspace of a vector space be non-empty. But does a vector space deserve to be called a set then?
This is yet published text which my professor uses for my classes. This class is Mathematical Methods/Numerical Analysis.
Empty set is a subset of every set can't be true. If a set S is a subset of a set V then there should be at least one element of set S that also belongs to a set V but there are no such element if the set S is empty. No?
I'm afraid the empty set is a subset of any set, at least in any (model of a) set theory worth its salt.
Well, this is it. The empty set is a subset of every set exactly because of the fact, that one doesn't need to verify, that every element of the empty set also belongs to a non-empty set. If we have a property which no elements of a non-empty set have, we obtain the empty subset of this non-empty set: [itex]\emptyset=\{x\in{M}|x\neq{x}\}[/itex] ...but as I eventually made clear for myself, it has nothing to do with a vector space...nothing can be defined on an empty set...from nothing comes nothing!
No, it's correct. [itex]x \not= x[/itex] is always false: there is no x which satisfies that nonequality, therefore the given set is empty. You didn't even specify what y is, so your expression has no meaning. To relieve any doubt, use the following definition of subset: If A and B are sets, then A is called a subset of B if: [tex]x \in A \Rightarrow x \in B[/tex] the notation is [itex]A \subset B[/itex]. Do you see now why the empty set is a subset of every set.
Also note that this property conveniently gives us the property that the intersection of two sets which have no elements in common is still a set, the empty set (since it is a subset of both sets).
Yeah, as others pointed out, the empty set can't be a vector space because it has no zero vector. However, the empty set does span the vector space consisting of the zero vector, according to the definition of span: The span of a set of vectors is the smallest subspace containing those vectors.
"The span of a set of vectors is the smallest subspace containing those vectors." :tongue2: Also, the zero vector is the linear combination of no vectors. (i.e. the "empty sum")
the summation symbol is a function from indexed sets of vectors to the vector space, which is additive on disjoint decompositions of the index set. hence it must send the empty index set (which is an indexed set, since the empty set is a set) to zero. hence the empty linear combination must equal zero. Many many proofs in linear algebra are lacunary without this remark. I just taught it last semester and found out how hard it is to ignore this case without making mistakes, (which however no one noticed except me). Indeed even the book I used made this mistake - although the authors tried hard to give correct statements without pointing out this subtle fact, they did not quite succeed. :rofl: