In his book on functional analysis Kreyszig gives the following definition:
Definition. A vector space $X$ is said to be finite dimensional if there is a positive integer $n$ such that $X$ contains a linearly independent set of $n$ vectors whereas any set of $n+1$ or more vectors of $X$ is linearly dependent. $n$ is called the dimension of $X$, written $n=\text{dim } X$. By definition, $X=\{0\}$ is finite dimensional and $\text{dim } X=0$. If $X$ is not finite dimensional, it is said to be infinite dimensional. If $\text{dim } X=n$, a linearly independent $n$-tuple of vectors of $X$ is called a basis for $X$.
He then proves the following basic theorem
Theorem. Let $X$ be an $n$ dimensional vector space. Then any proper subspace $Y$ of $X$ has dimension less than $n$.
Proof. If $n=0$, then $X=\{0\}$ and has no proper subspace. If $\text{dim } Y=0$, then $Y=\{0\}$, and $X \neq Y$ implies $\text{dim } X\geq 1$. Clearly, $\text{dim } Y\leq \text{dim } X=n$. If $\text{dim } Y$ were $n$, then $Y$ would have a basis of $n$ elements, which would also be a basis for $X$ since $\text{dim } X=n$, so that $X=Y$. This shows that any linearly independent set of vectors in $Y$ must have fewer than $n$ elements, and $\text{dim } Y<n$.
Questions:
- Why is the last sentence in the proof necessary? It seems like a contradiction has already been reached at this point.
- Am not sure the statement $\text{dim } Y\leq \text{dim } X$ is so clear given his definitions. We cannot have $\text{dim } Y=m>n$, but how can we rule out the case of $\text{dim } Y=\infty$?
We idea to rule out $\text{dim } Y=\infty$ is the following argument:
If $\text{dim } Y=\infty$, then in particular $\text{dim } Y\neq n$. Hence either there exist a set of $n+1$ linearly independent vectors in $Y$ or every set of $n$ vectors in $Y$ is linearly dependent. By definition of $\text{dim } X=n$ it must be that the latter holds. Now we repeat the argument to obtain that every set of $n-1$ vectors in $Y$ is linearly dependent. After $n$ steps we reach the conclusion that $Y=\{0\}$, contradiction.
Is this correct?
The last sentence appears to be there for emphasis.
The basic logic that establishes $\dim Y\leq\dim X$ also works for the case $\dim Y=\infty$: if $\dim Y>n$ then $\dim Y$ contains an independent set of size $n$. The fact that this also holds for $\dim Y=\infty$ doesn't follow immediately from the definition but it's not hard to prove. With $U$ a vector space, let $N$ be the set of all $k$ such that $U$ contains a set of $k$ independent vectors. $N$ is always non-empty since it contains $0$, and it is closed downwards (if $l<k$ and $k\in N$ then $l\in N$). The dimension of $U$ is just the maximum of $N$, or $\infty$ if the maximum doesn't exist. In either case it's clear that if $n<\dim U$, $n\in N$.
Your argument works but it's a little overkill since it stands on its own as a complete proof of the original theorem. To see this, notice that the only way you ever use the assumption that $\dim Y=\infty$ is to say that $\dim Y$ is not any of the numbers $0$ through $n$. So you're essentially proving by contradiction that $\dim Y\leq n$.