# Max dimension of subspace $U\subseteq V$, where $V$ is a vector space of functions $[0,1]\rightarrow \Bbb R$

Let $V$ be a vector space of functions $[0,1]\rightarrow \Bbb R$. What is the maximal possible dimension of $U\subseteq V$, a subspace consisting of monotone functions in $V$?

I was thinking to approach this question using elementary set theory which I learned to find the cardinality of $U$, but that wouldn't be very possible on the interval $[0,1]$.

I've never seen a question like this so I don't know what else can be done.

• After reading again your question, there is something strange. The subset of the monotone functions is not a vector subspace. So are you meaning that $U$ is the subspace spanned by the monotone functions. – mathcounterexamples.net Jun 21 '18 at 9:43
• @mathcounterexamples.net I suspect that what the OP has in mind is a vector space such that all of its elements are monotonic functions. – José Carlos Santos Jun 21 '18 at 10:59
• Good that the question has been reformulated in a proper way! By the way, this is a much less interesting question than the question of the dimension of the subspace spanned by the monotone functions. – mathcounterexamples.net Jun 21 '18 at 11:24

Hint: If $f$ and $g$ are monotonic functions and none of them is a multiple of the other one, then there are real numbers $\alpha$ and $\beta$ such that $\alpha f+\beta g$ is not monotonic.

The cardinality of the monotone real functions defined on $[0,1]$ is equal to $\mathfrak{c}$. See What is the cardinality of a set of all monotonic functions on a segment [0,1]?

Hence the dimension of $U$ is at most $\mathfrak{c}$. Now you'll be able to prove that the family of functions

$$f_\alpha(x)= \begin{cases} 0 & 0 \le x <\alpha\\ 1 & \alpha \le x \le 1 \end{cases}$$ for $\alpha \in (0,1)$ is linearly independent (consider the discontinuities of a linear combination). The cardinality of $(f_\alpha)_{\alpha \in (0,1)}$ is $\mathfrak c$. Hence, the dimension of $U$ (over $\mathbb R$) is equal to $\mathfrak{c}$ (while the dimension of $V$ is equal to $2^{\mathfrak c}$).

• @mathcounetrexamples.ne I think the OP is talking about a vector space of monotone functions. – Kavi Rama Murthy Jun 21 '18 at 9:11
• @KaviRamaMurthy Yes understand. And that is also what I'm speaking of... linear independence and so on is about vector space. – mathcounterexamples.net Jun 21 '18 at 9:13
• this is not a Q. in set theory. I might have caused confusion by mentioning cardinality. of course, cardinality is different than dimension. and i'm looking for the dimension and not cardinality – Jneven Jun 21 '18 at 11:05
• You say you understand, but you clearly missed the point to Kavi's comment. The span of the $f_\alpha$ contains non-monotone functions. So this seems totally irrelevant... – David C. Ullrich Jun 21 '18 at 14:45
• @DavidC.Ullrich David. It is irrelevant and I'll delete the answer. However, if you look at the first version of the question (that wasn't properly formulated), it was not so irrelevant. See by the way also comment from the OP Jneven above. – mathcounterexamples.net Jun 21 '18 at 14:54