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Sets of the form v+X

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
Hey!! 😊

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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
(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?
Hi mathmari !!

Yes.
Btw, your definition \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*} is the same as \begin{equation*}L:=\mathbb{R}\begin{pmatrix}1 \\ 2\end{pmatrix}=\left \{\lambda\begin{pmatrix}1 \\ 2 \end{pmatrix}\mid \lambda\in\mathbb R\right \}\end{equation*}
I think the latter is a bit easier to understand. 🤔

(i)
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*}
Is it correct to use everywhere $\iff$ ?
If I try to follow the argument in the reverse direction, it doesn't make much sense.
That is, we'd get as the first reverse step
\begin{align*}x\in v+L\implies x=(v+\ell)+\ell_1=v+(\ell+\ell_1) , \ \ell+\ell_1\in L \end{align*}
It's not exactly wrong because we can verify each reverse step with some extra thought, but I think it's better to make the argument for the reverse direction separately.
That is easier for the reader to understand, not to mention that it is easy to make mistakes otherwise. 🤔

(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?
Yep. That works. (Nod)

(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*}
The last reverse step looks suspicious
$$y\in (v+w)+X\implies y=(v+w)+x$$
This is not generally true. If y in a set that does not imply that it is equal to some specific element in that set. 🧐
The converse is true though: if y is a specific element in a set, then y is an element of that set.
What is missing, is that we only require that "there is an $x\in X$" for which it applies, which you left out.
Either way, again I think it is better to make the reverse argument separately to avoid mistakes, and to make it easier for the reader to understand and verify. 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
Btw, your definition \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*} is the same as \begin{equation*}L:=\mathbb{R}\begin{pmatrix}1 \\ 2\end{pmatrix}=\left \{\lambda\begin{pmatrix}1 \\ 2 \end{pmatrix}\mid \lambda\in\mathbb R\right \}\end{equation*}
I think the latter is a bit easier to understand. 🤔
Now I am confused. I understand your definition, but why is this the same with $\left \{\begin{pmatrix}x \\ y \end{pmatrix}\mid 2x-y=0\right \}$ ? How do we get the "minus" from your definition? :unsure:


If I try to follow the argument in the reverse direction, it doesn't make much sense.
That is, we'd get as the first reverse step
\begin{align*}x\in v+L\implies x=(v+\ell)+\ell_1=v+(\ell+\ell_1) , \ \ell+\ell_1\in L \end{align*}
It's not exactly wrong because we can verify each reverse step with some extra thought, but I think it's better to make the argument for the reverse direction separately.
That is easier for the reader to understand, not to mention that it is easy to make mistakes otherwise. 🤔
So,to show that $w+L \subseteq v+L$ we have :
\begin{align*}x\in w+L& \Rightarrow x=w+\ell_1 , \ \ell_1\in L \\ & \Rightarrow x=(v+\ell)+\ell_1=v+(\ell+\ell_1) , \ \ell+\ell_1\in L \\ & \Rightarrow x\in v+L\end{align*}
And to show that $v+L \subseteq w+L$ we have :
\begin{align*}x\in v+L& \Rightarrow x=v+\tilde{\ell} , \ \tilde{\ell}\in L \\ & \Rightarrow x=v+\ell-\ell+\tilde{\ell} , \ \tilde{\ell}\in L \\ & \Rightarrow x=w+(\tilde{\ell}-\ell) , \ \tilde{\ell}\in L \\ & \Rightarrow x\in w+ L\end{align*}

:unsure:


Yep. That works. (Nod)
I got stuck right now. Because at the previous statement we had $w$ and then the set $w+L$. Here we take an arbitrary $x$, or not? :unsure:


The last reverse step looks suspicious
$$y\in (v+w)+X\implies y=(v+w)+x$$
This is not generally true. If y in a set that does not imply that it is equal to some specific element in that set. 🧐
The converse is true though: if y is a specific element in a set, then y is an element of that set.
What is missing, is that we only require that "there is an $x\in X$" for which it applies, which you left out.
Either way, again I think it is better to make the reverse argument separately to avoid mistakes, and to make it easier for the reader to understand and verify. 🤔
It holds that
\begin{align*}y\in v+(w+X) &\Rightarrow y=v+(w+x) , \text{ with } x\in X \\ & \Rightarrow y=(v+w)+x \\ & \Rightarrow y\in (v+w)+X\end{align*}
It also holds that
\begin{align*}y\in (v+w)+X &\Rightarrow y=(v+w)+x , \text{ with } x\in X \\ & \Rightarrow y=v+(w+x) \\ & \Rightarrow y\in v+(w+X)\end{align*}
Sowe get the equality $v+(w+X)=(v+w)+X$, right? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Now I am confused. I understand your definition, but why is this the same with $\left \{\begin{pmatrix}x \\ y \end{pmatrix}\mid 2x-y=0\right \}$ ? How do we get the "minus" from your definition?
We get $x=\lambda$ and $y=2\lambda$ from "my" definition.
Substitute in $2x-y=0$ to find $2\cdot\lambda - 2\lambda=0$.
So they are the same aren't they? 🤔

Your version is actually a reworked version of the general definition, which is the one I gave.
Yours is a bit of an awkward definition though. (Bandit)

I got stuck right now. Because at the previous statement we had $w$ and then the set $w+L$. Here we take an arbitrary $x$, or not?
Didn't we have $w$ and then the set $v+L$ instead in the previous statement?
Now we have $x\in w+L\implies x+L=w+L$ when we apply the previous statement. And we also have $x\in v+L\implies x+L=v+L$. 🤔


The rest is all correct. (Nod)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
We get $x=\lambda$ and $y=2\lambda$ from "my" definition.
Substitute in $2x-y=0$ to find $2\cdot\lambda - 2\lambda=0$.
So they are the same aren't they? 🤔

Your version is actually a reworked version of the general definition, which is the one I gave.
Yours is a bit of an awkward definition though. (Bandit)



Didn't we have $w$ and then the set $v+L$ instead in the previous statement?
Now we have $x\in w+L\implies x+L=w+L$ when we apply the previous statement. And we also have $x\in v+L\implies x+L=v+L$. 🤔
Thanks for the explanation, I understodd that now!! :giggle:


As for the graphs.

We have that $$\begin{pmatrix}x\\ y\end{pmatrix}\in L_1 \ \text{ then } \ \begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}3\\ 0\end{pmatrix}+\begin{pmatrix}\lambda \\ 2\lambda \end{pmatrix}=\begin{pmatrix}\lambda+3 \\ 2\lambda \end{pmatrix}$$
So we set $x=\lambda+3$ and $y=2\lambda \Rightarrow \lambda=\frac{y}{2}$ then substituting this in the first equation we get $x=\frac{y}{2}+3 \Rightarrow y=2x-6$.


\begin{equation*}\begin{pmatrix}x\\ y\end{pmatrix}\in L_2 \ \text{ then } \ \begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}-1\\ -1\end{pmatrix}+\begin{pmatrix}\lambda \\ 2\lambda \end{pmatrix}=\begin{pmatrix}\lambda-1 \\ 2\lambda-1 \end{pmatrix}\end{equation*}
So $x=\lambda-1$ and $y=2\lambda -1\Rightarrow \lambda=\frac{y+1}{2}$ so we get $x=\frac{y+1}{2}-1 \Rightarrow y=2x+1$.

\begin{equation*}\begin{pmatrix}x\\ y\end{pmatrix}\in L_3 \ \text{ then } \ \begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 1\end{pmatrix}+\begin{pmatrix}\lambda \\ 2\lambda \end{pmatrix}=\begin{pmatrix}\lambda \\ 2\lambda+1 \end{pmatrix}\end{equation*}
So $x=\lambda$ and $y=2\lambda +1\Rightarrow y=2x +1$. (So we get the same line as the previous one?)


:unsure:
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
As for the graphs.

(So we get the same line as the previous one?)
We can do it a little bit easier.
From "my" definition it should be clear that $L$ is a line through the origin with directional vector $\begin{pmatrix}1\\2\end{pmatrix}$.
The others are the same line but through the points $\begin{pmatrix}3 \\ 0\end{pmatrix}$, $\begin{pmatrix}-1 \\ -1\end{pmatrix}$, respectively $\begin{pmatrix}0 \\ 1\end{pmatrix}$ instead of through the origin. :geek:

If we add the directional vector to the second one, we get the third one. So these are indeed the same lines. (Nod)