Recall what we covered in Lecture 10 - Finishing LI&LD, Bases#Bases. We made a system to show that the length of a LI list is $\le$ the length of a spanning list, by doing the "booting" process, where you boot out each $v$ one at a time by adding a $u$ each step:

Where since we eventually run out of $v$'s then the number of $u$'s is less than the number of $v$'s.

Infinite Dimensional Spaces

Let's get back into dimension. Recall that two unique bases in a vector space $V$, they have the same list length, which we the name of dimension. However, we might've jumped the gun, since we may not be finite dimensional. But first, what is that?

Note that the spanning list doesn't need to have the dimensions' number of vectors. We may have extras! But we can always trim it back down to a basis via Lecture 10 - Finishing LI&LD, Bases#Reduction.

For instance, $P(\mathbb{R})$, the space of all polynomials with real coefficients, is infinite-dimensional. But to prove it you prove that, for contradiction, if you had a finite spanning list for the space, then if you can create a vector not in that span that's still in $V$, then you did it! For our space in question, if we had a spanning list we'd have some maximum degree $x^n$, which doesn't include $x^{n+1}$, which completes the proof.

Let's do a cool proof!

Proof
We can't just use $V$'s spanning list to make a spanning list for $U$, since the vectors that are in that spanning list may not necessarily be in $U$. So we need to construct a new spanning list for it.

If $U=\{\overrightarrow{0}\}$, then the empty list $\{\}$ is the spanning list and we are done, as it is F.D.
If $U\ne \{\overrightarrow{0}\}$ then there is some vector $u_1\in U$. Consider $\text{span}(u_1)$. This list is LI already. If the span is $U$ itself then we are done. If it doesn't, then choose another vector $u_2\in U$ such that $u_2\notin \text{span}(u_1)$. If we consider $\text{span}(u_1,u_2)$, then this is LI, since we are adding a vector that isn't in the span of the previous list.

We repeat this process.

We always create a longer and longer LI list. But they can't be longer than the the spanning list for $V$, which is finite dimensional, say some dimension $n$. Then we must have the new spanning list at the end $\text{span}(u_1,...,u_i)$ must be smaller than $n$, so then $\text{dim}(U)\le \text{dim}(V)$. Furthermore, as a result we must terminate, so then $U$ is F.D.

Really, $\text{dim}(U)\le \text{dim}(V)$ comes from the Lecture 10 - Finishing LI&LD, Bases#Finishing 2.A, ie the Linear Independence Lemma, and $V$ being finite dimensional.
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Note that you have to construct this list, instead of using the list that spans $V$. Even though it is LI, the subset of vectors that are in $U$ may suddenly not be LI, which is the problem. So this process is needed.

There are also countably-infinite dimensional vector spaces like $P_n(\mathbb{R})$, where a bijection is made from each basis vector and $\mathbb{N}$. Furthermore, there's uncountably-infinite dimensional vector spaces, like the space of all functions ${\mathbb{R}}^{\mathbb{R}}$. However, the pretty results we get are based on being finite dimensional, so we'll focus on that until we get to the end of the quarter.

Proof
We need to prove that $U\cap W=\{0\}$.

Since $U$ is a subspace of finite-dimensional $V$, then $U$ is finite dimensional, so let $u_1,...,u_k$ be said basis for $U$. Now, extend this basis to a basis for $V$ via:

Where the $n-k$ is from $V$ being $n$ dimensional. This whole list has $k$ vectors. Consider just:

u_1, ..., u_k, v_1, ..., v_