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?

 

[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 V_n...

 

[yaa09d]

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