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Let $W$ be a subspace of a vector space $V$. Define relation $\sim$: $\mathbf{v}_1\sim\mathbf{v}_2$ if $\mathbf{v}_1-\mathbf{v}_2\in W$.

Denote by $\overline{\mathbf{v}}:=\{\mathbf{x}\in V\mid \mathbf{x}\sim \mathbf{v}\}$, the equivalence class containing $\mathbf{v}$. I have already shown that $\overline{\mathbf{v}} = \mathbf{v}+W$.

Now I want to proceed to show that the cardinality $|\overline{\mathbf{v}_\alpha}|=|\overline{\mathbf{v}_\beta}|$ for any $\alpha,\beta$.

I am thinking about showing the bijectivity of the map: $f:|\overline{\mathbf{v}_\alpha}|\mapsto|\overline{\mathbf{v}_\beta}|$. However, I have no clue how to proceed on other than I need to show both injection and surjection.

Could anyone hint me how to show such bijection? I am totally lost as it is my first time dealing with cardinality.

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    What is the map $f:|\overline{\mathbf{v}_\alpha}|\mapsto|\overline{\mathbf{v}_\beta}|$, the bijectivity of which you want to show? – 2017-01-16
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    By what you have already shown, isn't the cardinality of each of these equivalent classes equal to $|W|$? – 2017-01-16
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    Actually the hint of this question requires me to show the bijectivity of the map mentioned above, which I do not have any idea, in order to show the equality of cardinality. – 2017-01-16
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    I genuinely appreciate if you could explain a bit more. – 2017-01-16

2 Answers 2

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The function $$f:v_{\alpha}+W\to :v_{\beta}+W,\quad f(v_\alpha +w)=v_\beta +w $$ for all $w\in W$ satisfies $$f(v_\alpha +w)=(v_\alpha +w')\Rightarrow v_\beta +w=v_\beta +w'$$ $$\Rightarrow w=w'\Rightarrow v_\alpha +w=v_\alpha +w'\Rightarrow f\text{ is injective}.$$

$$\forall\; v_\beta +w\in v_{\beta}+W\Rightarrow f(v_\alpha +w)=v_\beta +w\Rightarrow f\text{ is surjective}.$$

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I guess you want to define a bijection: $f:|\overline{\mathbf{v}_\alpha}|\mapsto|\overline{\mathbf{v}_\beta}|$.

Just try $$f(w_\alpha)=f(v_\alpha+z_\alpha)=v_\beta+z_\alpha,\quad\forall \,w_\alpha=v_\alpha+z_\alpha\in\overline{v_\alpha}=v_\alpha+W.$$

Hope this helps.