Subsets and Subspaces

Date: 09/07/2001 at 01:29:23
From: Bernard Doria
Subject: Properties of linear span in set notation

I'm trying to extract some relationships in set notation from the 
following description from Loomis and Sternberg's _Advanced_Calculus_:

By Theorem 1.1 the linear span L(A) of an arbitrary subset A of a 
vector space V has the following two properties:
 i) L(A) is a subspace of V which includes A;
ii) If M is any subspace which includes A, then L(A) is a subset of M.

(By the way, Theorem 1.1 says that: If A is a nonempty subset of a 
vector space V, then the set L(A) of all linear combinations of the 
vectors in A is a subspace, and it is the smallest subspace of V 
which includes the set A.)

Here's what I gather:
L(A) is a subset of A.
A is a subset of V.
M is a subset of V.

Are these correct relationships? Are there any other relationships 
that can be derived from the description above?

Date: 09/07/2001 at 19:33:08
From: Doctor Roy
Subject: Re: Properties of linear span in set notation

Hello and thanks for writing to Dr. Math.

You are a little off. 

A is a subset in V.  This is true but not a relationship that is from 
the theorem, simply a given assumption. Likewise, M is a subspace of V 
(not a subset) that happens to contain all the elements of A. It isn't 
some special subspace, but some given hypothetical subspace.

L(A) is not a subset at all. It is a subspace. If you are confused 
over the distinction, a set is simply a collection of objects. A space 
has other properties (like an identity element, for example). In the 
theorem, it states that A is some arbitrary subset (not subspace).  
So, it is simply a collection of vectors from your space (not set) V.  
L(A) is a subspace (not subset) which contains A. So, any vector in A 
is contained in the subspace L(A). L(A) is the space formed by all 
linear combinations of vectors in A. 

The second part of the theorem is quite important.  It states that if 
there is a subspace M that contains A, then the subspace M also 
contains the subspace (not subset) L(A). 

I hope this helps, and feel free to write again.

- Doctor Roy, The Math Forum
  http://mathforum.org/dr.math/