# Inequality $\dim(F)\le K$ for all finite dimensional vector spaces of $E$ implies $\dim(E)\le K$

Let $E$ be a vector space. Suppose that for all finite dimensional vector subspaces $F\subset E$ we have $\dim(F)\le K$ where $K\in\Bbb{N}.$

How can I prove that $\dim(E)\le K?$

My try: Suppose that $E$ is infinite dimensional vector space, then there exist a Hamel Basis $(e_n)_n.$ I can extract a countable infinite linearly independent family $\{v_n; n\in \Bbb{N}\}.$

So if I take $F_1=span(v_1); F_2=span(v_1,v_2),\ldots,F_n=span(v_1,v_2,\ldots,v_n)$ I get that $\forall i\;F_i$ if finite dimensional and $F_i\subset F_{i+1}.$ Ok I have $\forall i\quad \dim(F_i)\le K.$

Not sure how can I continue.

• One should note that $K$ is actually a natural number and not an ordinal number. Also note that you can proof this without invoking Choice, so you should avoid doing so. I.e. do not use a Hamel basis please. – MooS Nov 4 '17 at 14:08
• @MooS thanks, but as you can see my level is not so high to talk about ordinal ;p. – Alex Nov 4 '17 at 14:10
• Let $F_0$ a subspace of $E$ of maximal finite dimension say $M$, we have $M\leq K$. Let $x\in E$. Can the intersection of the subspace $<x>$ generated by $x$ and $F_0$ be reduced to $\{0\}$ ? – Kelenner Nov 4 '17 at 14:13
• @Kelenner Hum. If it reduced to $\{0\}$ then $\dim(\langle x\rangle\cap F_0)=0$ so that $\dim(\langle x\rangle\cup F_0)=M+1\le K$ which is a(Edit :an)absurdity because $M$ is the maximal satisfying this property. The existence of $F_0$ is by Zorn's Lemma ? – Alex Nov 4 '17 at 14:22
• I think not. As there is only a finite number of possibilities for the dimension of the finite dimensional subspaces, call this set $A\subset \mathbb{N}$, you are taking the maximal term in this finite set $A$ of integers. – Kelenner Nov 4 '17 at 14:40