3

Let $S=\{0,2,4,6,8\}$ and $T=\{1,3,5,7\}$. Determine whether each of the following sets of ordered pairs is a function with domain $S$ and co-domain $T$.

  1. $\{(6,3),(2,1),(0,3),(8,7),(4,5)\}$
    TRUE
    This is a function
  2. $\{(2,1),(4,5),(6,3)\}$
    FALSE
    Not all domain values used
  3. $\{(0,2),(2,4),(4,6),(6,0),(8,2)\}$
    FALSE
    Domain values mapped to values outside of co-domain
  4. $\{(2,3),(4,7),(0,1),(6,5),(8,7)\}$
    TRUE
    This is a function
  5. $\{(6,1),(0,3),(4,1),(0,7),(2,5),(8,5)\}$
    FALSE
    $0$ is mapped twice

Currently, I am unsure about number 3, can the domain values be mapped to itself as shown or does that invalidate it as a function in this situation since the co-domain has specified values? Thanks!

Em.
  • 15,981
fluffy dog
  • 135
  • 4
    Looks good to me. – Alex Wertheim Jul 23 '16 at 06:32
  • You can use $\{1,2,3\}$ to show ${1,2,3}$. Please try not to use ALL CAPITAL LETTERS. I can be understood as shouting. Formatting tips here. – Em. Jul 23 '16 at 06:50
  • 2
    The only places it is allowed to be mapped to is the codomain. If the elements which are mapped to themselves happen to be elements of both the domain and codomain simultaneously, then there is no problem, but when those elements are only members of the domain and not the codomain, then you run into a problem describing it as a function from $S$ to $T$ as you would have hoped to do. Note that a function, $f$, from $S$ to $T$ is a relation from $S$ to $T$ (which satisfies some extra conditions), implying that every $(s,t)\in f$ satisfies $s\in S$ and $t\in T$. – JMoravitz Jul 23 '16 at 07:28
  • thanks! that clears my question up! – fluffy dog Jul 23 '16 at 08:19

1 Answers1

1

The domain values of a function must be mapped to values in the codomain, so, as indicated in the comments, your work is correct.

N. F. Taussig
  • 76,571