Onto and One-to-one proof

Question

I'm having a bit of difficulty with this. Where does the thinking come from? For equations it is pretty straight forward, but not for these abstract ones.

Let $D$ be the set of all infinite subsets of positive integers and define $T:\mathbb{Z}^+ \to D$ by the rule: $\forall n\in\mathbb{Z}^+, T(n) = \{d\in\mathbb{Z}^+: d\mid n\}$.

Is $T$ one-to-one, is it onto? Prove or give a counterexample.

I'm not sure how to find if it is one to one, but I think this might work for the onto bit?

If you have the set $\{1,2,3\}$. They are all divisors of $6$, but $6$ is also a divisor of $6$ but it is not in the set? I'm not really sure...

so continuing your remark, is ${1,2,3}$ in the image of $T$? — Dustan Levenstein, Nov 15 '13 at 06:15
The set of divisors of the positive integer $n$ is a finite set. Please look over your question and carefully edit. — André Nicolas, Nov 15 '13 at 06:16
I think the OP meant D is the set of all possible sets of positive integers. — GTX OC, Nov 15 '13 at 06:23
I am not able to guess. Maybe $T(n)$ is the set of multiples of $n$. — André Nicolas, Nov 15 '13 at 06:31

score 0 · Answer 1 · answered Nov 15 '13 at 06:15

The function as you stated isn't well defined. Maybe $D$ should be the set of all finites subsets of $\mathbb{N}$.

About one to one think of two different numbers, in each of their images the number is contained, can the images be the same?

About onto, may a number can have the image $\{6\}$? Could their be any number $n$ such that $1\notin T(n)$

Onto and One-to-one proof

1 Answers1