Any two right cosets of V/W are either disjoint or identical

Any two right cosets of V/W are either disjoint or identical

Quotient Space

    If V is a vector space over $\mathbb{R}$ and W is a subspace of V then $\frac{V}{W}=\{w+\alpha | \alpha\in V\}$ is quotient space and two operations addition and scalar multiplication on $\frac{V}{W}$ is defined as follows:
            Let $\alpha, \beta$ be any two arbitrary element of V then $W+\alpha, W+\beta\in \frac{V}{W}$ 
$(W+\alpha)+(W+\beta)=W+\alpha+\beta$ and 
$c(W+\alpha)=W+c\alpha$, for any $c\in\mathbb{R}$

Theorem :

Any two right cosets of $\frac{V}{W}$ are either disjoint or identical.

Proof :

Let $(W+\alpha)$ and $(W+\beta)$ be any two right cosets of W in V where $\alpha, \beta \in V$.
Claim : $(W+\alpha)\cap (W+\beta)=\phi$ or $(W+\alpha)=(W+\beta)$
Suppose, if possible, $(W+\alpha)\cap (W+\beta)\ne\phi$ 
Then, we have to prove that $(W+\alpha)=(W+\beta)$. 
As $(W+\alpha)\cap (W+\beta)\ne\phi$, there exist a vector $v\in V$ such that $v\in(W+\alpha)\cap (W+\beta)$. 
$\implies v\in W+\alpha$ and $v\in W+\beta$ 
As $v\in W+\alpha\implies \exists w_1\in W$ such that $v=w_1+\alpha$ 
Similarly, $v\in W+\beta\implies \exists w_2\in W$ such that $v=w_2+\beta$ 
$\therefore w_1+\alpha=w_2+\beta$ 
$\implies \alpha-\beta=w_2-w_1$ 
Since, $w_1, w_2\in W$ and W is a subspace of V, $\therefore w_2-w_1\in W$. 
$\therefore (\alpha-\beta)$ is also a vector in W. Let $u=\alpha-\beta$ be a vector in W. 
Now, to prove that (i) $W+\alpha\subseteq W+\beta$
                              (ii) $W+\beta\subseteq W+\alpha$ 
Let x be any vector in $W+\alpha$.
We will prove that $x\in W+\beta$.
Now, as $x\in W+\alpha\implies \exists w\in W$ such that,
$x=W+\alpha$ 
    $=w+(\alpha-\beta)+\beta$ 
    $=w+u+\beta$ 
    $=w'+\beta$ 
$\implies x\in W+\beta$ 
$\therefore W+\alpha\subseteq W+\beta$ 
Let $y\in W+\beta\implies y=w'_1+\beta, w'_1\in W$ 
                                                $=w'_1+w''_1+\alpha\in W+\alpha$ 
$\therefore W+\beta\subseteq W+\alpha$ 
$W+\alpha=W+\beta$ 
Hence proved.

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