# A calculus of the absurd

##### 22.3.4 Subspaces
• Definition 22.3.3 Let $$\textsf {V}$$ be a vector space. The set $$\textsf {W}$$ is a subspace of $$\textsf {V}$$ if $$\textsf {W}$$ is a vector space, and $$\textsf {W}$$ is a subset of $$V$$.

• Technique 22.3.1 Showing that something is a subspace. Suppose we have a vector space $$\textsf {V}$$, and we want to prove that $$\textsf {W}$$ is a subspace of $$\textsf {V}$$. The steps to do so are this

• 1. Show that the zero vector is in the subspace in question.

• 2. Show that $$W \subseteq V$$ using the standard technique for showing that something is a subset of something else (as in Section TODO: write).

• 3. Then we must show that $$W$$ is closed under vector addition and scalar multiplication. The rest of the vector space axioms follow from the fact that $$W \subseteq V$$ and $$V$$ is a vector space.

This theorem is given as both an example of how to prove facts about vector spaces, but also because it is important in its own right.

• Theorem 22.3.1 Let $$\textsf {V}$$ be a vector space, and $$\textsf {U}$$ and $$\textsf {W}$$ be subspaces of $$\textsf {V}$$. Prove that $$U \cup W$$ is a subspace of $$V$$ if and only if $$U \subseteq W$$ or $$W \subseteq U$$.

To prove this, first we will show the “if” direction, and then the only if direction.

• If. Without loss of generality, assume that $$U \subseteq W$$, in which case $$U \cup W = W$$, and this is a subspace of $$V$$ as $$W$$ is a subspace of $$V$$. The proof for the other case follows by swapping $$U$$ and $$W$$ in the proof.

• Only if. This direction requires a bit more of an intuition about what directions to explore. First we will assume that $$U \cup W$$ is a subspace, and then we will assume that the consequent is untrue (i.e. that $$U \subseteq W \text { or } W \subseteq U$$ is not true), in which case there exist $$u, w \in \textsf {V}$$ such that

$$u \in U \setminus W \text { and } w \in W \setminus U. \label {def of u and w}$$

We can then ask (this is the core idea in the proof which is not immediately obvious – to me at least), about the status of $$u + w$$. As $$u, w \in U \cup W$$ and by assumption $$U \cup W$$ is a subspace (and therefore by definition closed under vector addition) it must be that $$u + w \in U \cup W$$. Then either $$u + w \in U$$ or $$u + w \in W$$ (by definition of the set union).

• 1. If $$u + w \in U$$, then also $$u + w + (-u) = w$$ which is a contradiction as by definition of $$w$$ (Equation ??) $$w \notin U$$.

• 2. If $$u + w \in W$$, a very similar thing is the case; also $$u + w + (-w) = u \in W$$ which is a contradiction as $$u \notin W$$.

Therefore, by contradiction this direction of the theorem must be true.