- #1

mathmari

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Let $v\in \mathbb{R}^2$ be a vector and let $X\subseteq \mathbb{R}^2$ be a subset, then we define the subset of $\mathbb{R}^2$ : $$v+X:=\{v+x\mid x\in X\}$$

Let \begin{equation*}L:=\mathbb{R}\begin{pmatrix}1 \\ 2\end{pmatrix}=\left \{\begin{pmatrix}x \\ y \end{pmatrix}\mid 2x-y=0\right \}\end{equation*} and let \begin{equation*}L_1:=\begin{pmatrix}3 \\ 0\end{pmatrix}+L , \ , L_2:=\begin{pmatrix}-1 \\ -1\end{pmatrix}+L , \ , L_3:=\begin{pmatrix}0 \\ 1\end{pmatrix}+L\end{equation*}

(a) Draw the sets $L, L_1, L_2, L_3$ in a coordinate system and show the following for each $v, w\in \mathbb{R}^2$ :

(i) $w\in v+L\Rightarrow w+L=v+L$

(ii) $(v+L)\cap (w+L)\neq \emptyset \Rightarrow w+L=v+L$

(b) Show for all $v, w\in \mathbb{R}^2$ and $X\subseteq \mathbb{R}^2$ that $v+(w+X)=(v+w)+X$.

I have done the following:

(a) To draw the sets : $L$is the points on the line $y=2x$, right? All the other sets are lines that have been shifted?

(i) Let $w\in v+L$.

This means that $w=v+\ell$, with $\ell\in L$.

We have that \begin{align*}x\in w+L& \iff x=w+\ell_1 , \ \ell_1\in L \\ & \iff x=(v+\ell)+\ell_1=v+(\ell+\ell_1) , \ \ell+\ell_1\in L \\ & \iff x\in v+L\end{align*}

So we get the equality $w+L=v+L$.

Is it corrct to use everywhere $\iff$ ?

(ii) Let $(v+L)\cap (w+L)\neq \emptyset$.

Then there is an element in the intersection, say $x$.

So $x\in (v+L)\cap (w+L)$ means that $x\in (v+L)$ and $x\in (w+L)$. Do we use now the stetement of (i) to get the desired result? (b) Do we have here the following equivalences \begin{align*}y\in v+(w+X) &\iff y=v+(w+x) , \text{ with } x\in X \\ & \iff y=(v+w)+x \\ & \iff y\in (v+w)+X\end{align*} :unsure: