Cardinality of a vector space over an infinite field

[yaa09d]

Let V be a vector space over an infinite field $\mathbf{k}$. Let \beta be a basis of V.

In this case we can write

V\cong \mathbf{k}^{\oplus \beta}:=\bigl\{ f\colon\beta\to \mathbf{k}\bigm| f(\mathbf{b})=\mathbf{0}\text{ for all but finitely many }\mathbf{b}\in\beta\bigr\}.


Q:Show that card(V) = card( \mathbf{k} ) card(\beta )
Can anyone help?:smile:

[micromass]

Let

V_n=\{f:\beta\rightarrow k~|~f(b)=0~\text{except for possibly n values of b}\}

It is clear that |V_n|=|\beta||k|. Then

V\cong \bigcup_{n\in \mathbb{N}}{V_n}. Thus |V|=|\beta||k|...

[yaa09d]

It is clear that |V_n|=|\beta||k|.

Thank you for your quick reply, but how is that clear?

[micromass]

Well, it isn't that clear, but you should think about it. The following would probably make it easier:

Take V1. Then to construct a map in V1, then you just need to select an element b in \beta and x in k. Then the map is defined by f(b)=x and all other elements map to 0. Thus |V_1|=|k||\beta|.

Take V2. Then to construct a map in V2, then you just need to select elements b, b' in \beta and x,y in k. Then define a map by f(b)=x and f(b')=y and all other elements map to 0. Thus |V_2|=|k|^2|\beta|^2=|k||\beta| since k is infinite.

The same happens with the other Vn...

[yaa09d]

I see. It's clear now. Thank you.