1. ## Aother subset proof

Show that if S1 and S2 are arbitrary subsets of a vector space V, then span(S1 union S2) = Span(S1)+Span(S2).
Can someone solve this?

And did I approach this correctly

1. Let s1 and s2 be arbitrary subsets of a v. space V.
2. Let x be an arbitrary vector such that x belongs to (S1 union S2). So x belongs to S1 or x belongs to S2. Then Span(S1) IS all the linear combinations containing the vector x such that (for all a belonging to the field F)(a is a scalar) then the summation a*x belongs to Span(S1)
and then.....

my wordin prob sounds bad i dont know how u guys use the summation symbol and the for all symbol on the internet... how do u guys do that anyhow

2. Originally Posted by ruprotein
Show that if S1 and S2 are arbitrary subsets of a vector space V, then span(S1 union S2) = Span(S1)+Span(S2).
Can someone solve this?
$\mbox{span}(S_1\cup S_2)\not = \mbox{span}S_1+\mbox{span}S_2$
Because the right hand side is the set of ordered paris of $S_1$ and $S_2$.

Hence your probably mean to say (and you tell me what):
1) $\mbox{span}(S_1\cup S_2)=\mbox{span}(S_1)\cup \mbox{span}(S_2)$
2) $\mbox{span}(S_1\cup S_2)\simeq \mbox{span}(S_1)+\mbox{span}(S_2)$.

By $\simeq$ I mean, isomorphic.

3. Actually he does mean $span\left( {S_1 \cup S_2 } \right) = span\left( {S_1 } \right) + span\left( {S_2 } \right).$ The term $\left( {S_1 \cup S_2 } \right)$ is only a union not a set of ordered pairs.

As for a proof, it is easy to see that $span\left( {S_1 } \right) + span\left( {S_2 } \right) \subseteq span\left( {S_1 \cup S_2 } \right)$.
To prove $span\left( {S_1 \cup S_2 } \right) \subseteq span\left( {S_1 } \right) + span\left( {S_2 } \right)$ is just a messy exercise. In any linear combination from $span\left( {S_1 \cup S_2 } \right)$ you need to separate the spanning elements into one of those two sets.