Linear Algebra with Proof by Contradiction

[Devil Moo]

This is a linear algebra question which I am confused.

1. Homework Statement

Prove that "if the union of two subspaces of $V$ is a subspace of $V$, then one of the subspaces is contained in the other".

The Attempt at a Solution

Suppose $U$, $W$ are subspaces of $V$. $U \cup W$ is a subspace of $V$. (statement A)

Suppose $U$ is not contained in $W$ and vice versa. (statement B)

Let $u \in U, \not\in W$ and $w \in W, \not\in U$.

$u \in U \cup W, w \in U \cup W$
$u + w \in U \cup W$
$u + w \in U$ or $u + w \in W$

Suppose $u + w \in U$ (statement C)

$u + w + (-u) \in U$
$w \in U$
It leads to contradiction.

Which one does it conclude? (if A is true, then B is false) or (if A and B are true, then C is false)

 

[andrewkirk]

It's the second one: (if A and B are true, then C is false)

Since you have proven C false, you can conclude that $u+v\in W$. By similar steps to before you can also conclude that that is false.

You can then, with a few more steps, conclude that at least one of A or B must be false.

Arguments of this kind are much easier to understand and control if you grasp the notion of Conditional Proof or sub-proof. Every time you make a new assumption, you are opening a new sub-proof. When you reach a contradiction, you close that sub-proof, concluding the opposite of the assumption that you used to open it. It is not unusual to have several nested levels of proof. In your case there are three levels.

 

[Devil Moo]

For the proof by contradiction about $\sqrt 2$ is a irrational number, we conclude that it is true once we find the contradiction.
In this case, why can't I conclude that $u + v \in W$ is true?

Is $not (u + v \in U) = u + v \in W$ wrong?

 

[andrewkirk]

In this case, why can't I conclude that $u + v \in W$ is true?

You can. Read the second line of my post again.

Then you go on to prove that it is also false, which is a second contradiction, which tells you that either A or B must be false.