Cardinality of a vector space over an infinite field

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Discussion Overview

The discussion centers on the cardinality of a vector space over an infinite field, specifically examining the relationship between the cardinality of the vector space \( V \), the field \( \mathbf{k} \), and the basis \( \beta \). The scope includes mathematical reasoning and exploration of concepts related to vector spaces and their dimensions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant states that for a vector space \( V \) over an infinite field \( \mathbf{k} \) with basis \( \beta \), the cardinality can be expressed as \( \text{card}(V) = \text{card}(\mathbf{k}) \cdot \text{card}(\beta) \).
  • Another participant introduces the concept of \( V_n \), which consists of functions from \( \beta \) to \( \mathbf{k} \) that are non-zero for at most \( n \) values, asserting that \( |V_n| = |\beta| \cdot |\mathbf{k}| \).
  • A participant questions the clarity of the assertion that \( |V_n| = |\beta| \cdot |\mathbf{k}| \), prompting further explanation.
  • In response, a participant elaborates on constructing maps in \( V_1 \) and \( V_2 \), explaining how to define these maps and confirming that \( |V_1| = |\mathbf{k}| \cdot |\beta| \) and \( |V_2| = |\mathbf{k}|^2 \cdot |\beta|^2 = |\mathbf{k}| \cdot |\beta| \) due to the infinite nature of \( \mathbf{k} \).
  • Another participant expresses understanding after the explanation, indicating that the reasoning is now clear.

Areas of Agreement / Disagreement

Participants appear to agree on the mathematical reasoning presented, but there is initial uncertainty regarding the clarity of the cardinality claims, which is addressed through further explanation.

Contextual Notes

The discussion involves assumptions about the properties of infinite fields and the nature of vector spaces, which may not be explicitly stated. The reasoning relies on the definitions of cardinality and the behavior of functions defined on bases of vector spaces.

yaa09d
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Let [tex]V[/tex] be a vector space over an infinite field [tex]$\mathbf{k}$[/tex]. Let [tex]\beta[/tex] be a basis of [tex]V[/tex].

In this case we can write

[tex]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\}.[/tex]

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

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

It is clear that [tex]|V_n|=|\beta||k|[/tex]. Then

[tex]V\cong \bigcup_{n\in \mathbb{N}}{V_n}[/tex]. Thus [tex]|V|=|\beta||k|[/tex]...
 
micromass said:
It is clear that [tex]|V_n|=|\beta||k|[/tex].

Thank you for your quick reply, but how is that clear?
 
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 [tex]\beta[/tex] and x in k. Then the map is defined by f(b)=x and all other elements map to 0. Thus [tex]|V_1|=|k||\beta|[/tex].

Take V2. Then to construct a map in V2, then you just need to select elements b, b' in [tex]\beta[/tex] 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 [tex]|V_2|=|k|^2|\beta|^2=|k||\beta|[/tex] since k is infinite.

The same happens with the other Vn...
 
I see. It's clear now. Thank you.
 

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