Let $S$ be the vector space of real sequences, and for $x=(x_1,x_2,\dots)$ define $\alpha(x)=(0,x_1,x_2,\dots)$ and $\beta(x)=(x_2,x_3,\dots)$. The problem was asking for few other things to do, but I got stuck at showing that the first is not injective, while the second is not surjective.
Now, I realize I need to find two distinct arguments (two different sequences), plug it in $\alpha(x)$ and get the same value, which would show that is not injective. But I can not think of any example. Could somebody guide me, how I should structure my search for such example?
As for second, could I take for example a sequence of one member $\{x_1\}$? Then it would be undefined for $\beta$ proving it's not surjective, right?
Best Answer
Maybe this is backwards? $\alpha(x)$ looks injective to me and $\beta(x)$ looks surjective.
$\alpha(x)$ is not surjective though, and I think you have the right idea. Basically, the realization is that $\alpha(x)$ has a $0$ as its first coordinate for each $x$, and so in particular, there is no sequence that maps to $(1, 1, 1,...)$.
$\beta(x)$ is not injective. Intuitively, this is because $\beta(x)$ does not give you enough information to reconstruct $x$. In particular, you have no idea what the first coordinate of $x$ is given $\beta(x)$. In fact, you can make this into a proof by picking $x, x'$ that differ only in the first coordinate and showing they map to the same thing.