I could use some help with this problem,
Suppose that V is a vector space over any field F and that U and W are subspaces of V.
Show that the only time the union of U and W is a subspace of V is when either U is a subset of W or W is a subset of U.
I could use some help with this problem,
Suppose that V is a vector space over any field F and that U and W are subspaces of V.
Show that the only time the union of U and W is a subspace of V is when either U is a subset of W or W is a subset of U.
Suppose that neither is a subset of the other.
This means $\displaystyle a \in U\backslash V\quad \& \quad b \in V\backslash U.$
If $\displaystyle U \cup V$ is a subspace then $\displaystyle a + b = c \in \left( {U \cup V} \right).$
But if $\displaystyle c \in U$ then $\displaystyle c - a \in U$ or $\displaystyle b \in U.$
That is a contradiction. Similarly if $\displaystyle c \in V$ we get a contradiction.
Thus one of U or V is a subset of the other.
Clearly, if one of U or V is a subset of the other then their union is a subspace.
Suppose that neither is a subset of the other.
This means
What is meat here is that there is an element 'a' which is in U but not in V
Similarly there is an element 'b' which is in V but not in U
This statement follows directly from the fact that U is not a subset of V and V is not a subset of U
If is a subspace then
both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.
But if then or
That is a contradiction. Similarly if we get a contradiction.
Thus one of U or V is a subset of the other.
As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!
Same can be argued for V.
Hence the result. Thanks
Suppose that neither is a subset of the other.
This means
What is meant here is that there is an element 'a' which is in U but not in V
Similarly there is an element 'b' which is in V but not in U
This statement follows directly from the fact that U is not a subset of V and V is not a subset of U
If is a subspace then
both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.
But if then or
That is a contradiction. Similarly if we get a contradiction.
Thus one of U or V is a subset of the other.
As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!
Same can be argued for V.
Hence the result. Thanks
Suppose that neither is a subset of the other.
This means
What is meant here is that there is an element 'a' which is in U but not in V
Similarly there is an element 'b' which is in V but not in U
This statement follows directly from the fact that U is not a subset of V and V is not a subset of U
If is a subspace then
both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.
But if then or
That is a contradiction. Similarly if we get a contradiction.
Thus one of U or V is a subset of the other.
As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!
Same can be argued for V.
Hence the result. Thanks
I realize it has been a while, but I was thinking about the same problem trying to prove that it either has to be U is a subset of V or V is a subset of U. However, I am confused. Are a,b,c, vectors?
Also, could you explain a little more about how you got to a+b=c and how that actually gives us the desired result.
Thanks!
$\displaystyle a, b$ and $\displaystyle c$ are vectors. If you assume that neither is a subspace of the other one, then there exist $\displaystyle a$ and $\displaystyle b$ as described above.
Since all live in a vector space, you can add $\displaystyle a$ and $\displaystyle b$, call the result c, so define $\displaystyle c=a+b$.
Now we assume that the union of $\displaystyle U$ and $\displaystyle V$ is a subspace. Then since $\displaystyle a,b \in (U \cup V)$, their sum (that is c) is also inside $\displaystyle U \cup V$.
Clear?