Mapping: one-to-one and onto

• Nov 6th 2008, 03:06 PM
dolphinlover
Mapping: one-to-one and onto
Hello all!

I went through this and thought I had a handle on it till I spoke with a classmate of mine. Hopefully someone can set me straight.

a.) Is there a 1-to-1 function from the set {1,2,3} to the set {1,3}? Why or why not?

b.) Is there a function mapping {1,2,3} onto the set {1,3}? Why or why not?

c.) Is there a 1-to-1 function mapping the open interval (0,2) to the open interval (0,1)?

I answered no to (a) yes to (b) and no to (c)....my classmate crossed the sets in part (a) and has answered yes.....Why would you cross them?

Thanks!!!!!!!
• Nov 6th 2008, 05:37 PM
Aryth
For a, you cannot have a 1-to-1 function because if we make:

$D = \{1,2,3 \}$

$C = \{1,3 \}$

$D \ x \ C = \{(1,1), (1,3), (2,1), (2,3), (3,1), (3,3) \}$

There are three elements of D and only two in C, A function:

$f : D \to C$

Cannot possibly be 1-to-1 because there is a third element in the domain that cannot go with an element of the range.

You were right with a.
• Nov 6th 2008, 05:45 PM
vincisonfire
I'm not sure but since intervals (0,1) and (0,2) have the same cardinality there must be an injection from one to another
• Nov 6th 2008, 07:02 PM
dolphinlover
Quote:

Originally Posted by vincisonfire
I'm not sure but since intervals (0,1) and (0,2) have the same cardinality there must be an injection from one to another

I'm not familiar with the mathematical term "injection."
• Nov 6th 2008, 07:13 PM
Aryth
It's saying that because the two intervals have an equal number of elements (known as equinumerous), there must be a function that exists that is a one-to-one function.
• Nov 6th 2008, 07:26 PM
dolphinlover
Quote:

Originally Posted by dolphinlover
Hello all!

I went through this and thought I had a handle on it till I spoke with a classmate of mine. Hopefully someone can set me straight.

a.) Is there a 1-to-1 function from the set {1,2,3} to the set {1,3}? Why or why not?

b.) Is there a function mapping {1,2,3} onto the set {1,3}? Why or why not?

c.) Is there a 1-to-1 function mapping the open interval (0,2) to the open interval (0,1)?

I answered no to (a) yes to (b) and no to (c)....my classmate crossed the sets in part (a) and has answered yes.....Why would you cross them?

Thanks!!!!!!!

I still don't get part (c)... Am I correct in thinking of the open interval (0,2) as being all real #'s between 0 & 2 (but not including)?
• Nov 6th 2008, 07:36 PM
Aryth
Yes, you think of the intervals that way but consider this:

The interval (0,1) will be the set of real numbers between 0 and 1, it can be shown that this will be a set of real numbers that is infinite and uncountable.

The interval (0,2) will be the set of real numbers between 0 and 2, it can also be shown that this will be a set of real numbers that is infinite and uncountable.

It can then be said that the two sets have the same number of elements and therefore a function must exist such that the function is one to one.

Since the two sets are equinumerous, then there is a function that MUST be bijective, meaning that there is a function that is one-to-one AND onto.

It's hard to understand but infinite is infinite, no matter how you divide it or try to dissect it.
• Nov 6th 2008, 07:41 PM
dolphinlover
Thank You! Our professor has told us we would be talking about sizes of infinity, but we haven't gotten to that yet. Thanks Again!