Suppose that $V_1$ and $V_2$ are subsets of a vector space... Suppose that $V_1$ and $V_2$ are subsets of a vector space, is $span(V_1\cup V_2) = span(V_1)\cup span(V_2)$?
This seems like it should be pretty straight-forward but something is baking my noodle. It seems that $span(V_1)\cup span(V_2)$ would include vectors that don't exist in either $span(V_1)$ or $span(V_2)$. Is this the case or am I missing something important about spanning?
The follow-up is very similar to the first part:
Is $span(V_1\cap V_2) = span(V_1)\cap span(V_2)$?
Again, my initial thought is no due to the same reasons as the first part.
I hope that is clear enough. Thank you for your help.
 A: Consider two linearly independent lines in $\Bbb R^2$ say. As the comment mentions, if $S,T$ are subspaces of a vector space $S\cup T$ is a subspace if and only if $S\subset T$ or $T\subset S$. Can you prove this?
A: A fact that many books don't contain is that $ span(W_1 \cup W_2) = W_1 + W_2 $ where $W_1 + W_2 = \{\alpha \ | \ \alpha  = \beta  + \gamma \ \  \text{for some $ \beta \in W_1 $ and $ \gamma \in W_2 $ } \} $. This should be quite easy to prove. Since every vector in $ span (W_1 \cup W_2) $ is a linear combination of vectors in $W_1$ and $W_2$ and as soon as you separate the linear combinations into a sum of two vectors from $W_1$ and $W_2$, they belong to $W_1 + W_2$. The other way too is quite obvious. 
As it has been mentioned above $span W_1 \cup span W_2$ is a vector space $\iff $ either $span W_1 \subseteq span W_2$ or $ span W_2 \subseteq span W_1 $. Note that in each case your result will be true. 
Second one need not be true either. Consider $\Bbb R^3$. $V_1 = \{(1,1,0)\}$ and $ V_2=\{(0,1,1)\} $. Then $V_1 \cap V_2 = \emptyset$ and hence its span too is empty. But $span V_1 \cap span V_2 = \{t(0,1,0) \ | \ t \in \Bbb R\}$. 
A: You are right that the answer is no for both cases, but the reason is not quite the same both times.
For the case of unions, as you observed the span of the union may contain vectors that are not in the union of the spans. Indeed a linear combination of vectors in $V_1\cup V_2$ will only obviously belong to the union of the spans of $V_1$ and of $V_2$ if either its coefficients of the elements of $V_2$ are all zero (so that it belongs to the span of $V_1$) or its coefficients of the elements of $V_1$ are all zero (so that it belongs to the span of $V_2$), and there is no reason why one of these special conditions should always hold.
This argument does not yet give a proof that the span of the union contains vectors that are not in the union of the spans, since linear combination giving a specified vector are not always unique: a vector given by a linear combination not satisfying either special condition (of having all zeros on some $V_i$) might still belong to the union of the spans of $V_1$ and of $V_2$ in a non-obvious way, if the linear combination can be replaced by another one that does meet a special condition. But it is easy to construct an example where linear combinations are unique (this happens when all elements of $V_1$ and $V_2$ linearly independent), and then any linear combination with nonzero coefficients both of elements of $V_1$ and of $V_2$ gives such a vector.
For the case of intersections the situation is different, because the span of the intersection could be way smaller than the intersection of the spans. The point is that knowing a span of a set does not at all allow reconstructing the set: many, many candidates (generating sets) for spanning a given subspace can in general be found. So even in the simple case where $V_1$ and $V_2$ each span the whole space (which is then also the intersection of their spans) there is no reason that $V_1$ and $V_2$ should even have a single element in common, and if not the span of their (empty) intersection is just the zero dimensional subspace.
