I am asked to find a set of vectors in $\mathbb{R}^2$ such that if $y$ is in the set, then $b\cdot y$ is in the set for every real number $b$. However, I am told that this set cannot be a vector space.
The set would not be a vector space if the zero vector weren't included. However, given that $b$ can be any real number, including $0$, this cannot be the case. How else can a set of vectors that matches this condition not be a vector space?
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Consider the set $S$ formed of $x=(a,c) \in \mathbb{R}^2$ such as $\left|a\right|\geq \left|c\right|$. Then for $b \in \mathbb{R}$ $$ bx=\left(ba,bc\right) $$ and $\left|ba\right| \geq \left|bc\right|$.
However, with $(4,1)$ and $(-3,2)$ which are both in the set $$ (4,1)+(-3,2)=(1,3) $$ which is not in the set so it is not a vector space.
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How else can a set of vectors that matches this condition not be a vector space?
Recall that there is another condition: besides being non-empty (or containing the zero vector) and being closed under scalar multiplication, it also needs to be closed under addition.
Hint: think of the union of the coordinate axes; i.e. the $x$- and $y$-axis.
Consider the set $S = X \cup Y$ with: $$\color{blue}{X = \left\{ (a,0) \in \mathbb{R}^2 \;\vert\; a \in \mathbb{R} \right\}}\quad,\quad \color{red}{Y = \left\{ (0,b) \in \mathbb{R}^2 \;\vert\; b \in \mathbb{R} \right\}}$$ Now verify that taking any $\vec s \in S$, you have for all $\lambda \in \mathbb{R}$ that $\lambda\vec s \in S$.
However, take $\color{blue}{\vec x} \in X$ and $\color{red}{\vec y} \in Y$ with $\vec x \ne \vec 0$ and $\vec y \ne \vec 0$; then consider $\color{purple}{\vec x + \vec y}$.
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Note that the corresponding space created by a vector(non-zero) is equivalent to a line passing through zero. Now since your assumption does not involve being closed under addition you only need to consider at least two lines which pass through zero.
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Choose any two linearly independent vectors $v,w$. Then the set
$$\{bv\mid b\in\Bbb R\}\cup\{bw\mid b\in\Bbb R\}.$$
satisfies your condition, but is not a vector space since $v+w$ is not contained (vector spaces must contain all sums of their vectors).
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You could just take the empty set. Since it is empty it isn't a vector space, but it is vacuously closed under scalar multiplication.
Obviously this is a somewhat cheap example, but it does solve the problem as stated, and underscores that existential properties (e.g. the existence of a 0 vector) can't be derived from universally quantified implications.
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Let $S$ be the union of the first and third quadrants (along with the $x$- and $y$-axes, let's say). Scaling by a real number $r$ will either keep each vector in its same quadrant (if $r>0$), or send the first and third quadrants to each other (if $r<0$), or send all vectors to the origin (if $r=0$). So $S$ is closed under scalar multiplication.
However, notice that $$ (-2, -1) + (1, 2) = (-1, 1) $$ which is in the second quadrant, so $S$ is not closed under vector addition.
Extra note: for an example of something closed under addition but not under scalar multiplication, take just the first quadrant.
Take the set $V:=\{v\in\mathbb{R}^2: v=(x,0), v=(0,y), x,y\in\mathbb{R}\}$ Clearly if $v\in V$ then for any $\alpha\in\mathbb{R}$ we have $\alpha v\in V$, but $v_1+v_2\notin V$ where $v_1=(x,0),v_2=(0,y)$ for any such pairs and $x\neq0, y\neq 0$.