Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
You got it mixed up. The first one is not surjective, while the second one is not injective. –  Julian Kuelshammer Jun 22 '13 at 14:30
    
Are you sure that you have to prove that $\alpha$ is not injective and $\beta$ is not surjective and not the other way around. I guess that $\alpha$ is not surjective because the sequence $(x_1,x_2,...)$ with $x_1 \neq 0$ isn't part of the image of $\alpha$. –  mvcouwen Jun 22 '13 at 14:33
    
lol, yeah, now I see. the book mixed me, but wasn't able to spot it. But then, I was wondering, I couldn't take sequence of one membe $\{x_1\}$ and plug it in $\beta$ showing its not surjective? –  Sarunas Jun 22 '13 at 14:33
    
..and the first one is injective while the second is surjective. These are standard examples. –  DonAntonio Jun 22 '13 at 14:34

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer

Hints:

$$\{1,1,...,1,\ldots\}\notin\text{Im}\,(\alpha)\,,\,\,\text{but}\;\;\{0,x_1,x_2,\ldots\}=\{0,y_1,\ldots\}\iff$$

$$ \{x_1,x_2,\ldots\}=\{y_1,y_2,\ldots\}$$

$$\forall\;x:=\{x_1,x_2,...\}\;,\;\;\beta\{0,x_1,x_2,\ldots\}=x\;,\;\;\text{yet}$$

$$\beta\{1,2,3,\ldots\}=\beta\{0,2,3,\ldots\}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.