Prove that $E(a) = E(b)$ for $a,b \in V$ iff $a - b$ is in a subspace of V

Question

Let $H$ be a subspace of $V$.

For $c \in V$, define $E(c) = \{c + h\,|\,h\in H\}$.

Show that for $a,b \in V, E(a) = E(b)$ iff $a - b\in H$.

My attempt: \begin{align*} E(a) &= \{a + h\,|\,h\in H\}\\ E(b) &= \{b + h\,|\,h\in H\} \end{align*} Since $a - b \in H$, \begin{align*} E(a) &= \{a + (a-b)\,|\,a-b\in H\}\\ &=\{2a - b\,|\,a-b\in H\}\\\\ E(b) &= \{b + (a-b)\,|\,a-b\in H\}\\ &= \{a\,|\,a-b\in H\} \end{align*}

Which does not show that $E(a) = E(b)$. How is it done?

Answer 1

I denote , commonly, $E(a)$ as $a+H$. So your question is : to prove $a+H=b+H \iff a-b \in H$

$\implies$: Assume $a+H=b+H$.

Now $$a=a+0 \in a+H=b+H$$ so that $a=b+h$ where $h \in H$. Thus, $a-b=h \in H$

$\impliedby:$ Assume $a-b \in H$

Let $k \in a+H$. Then $k=a+h$ where $h \in H$.

Now, $$k=a+h=b+\color{red}{(a-b)+h} \in b+H$$

Thus, $a+H \subseteq b+H$. Similarly prove other inclusion