• July 5th 2010, 11:05 AM
rainer

Thanks
• July 5th 2010, 12:04 PM
HallsofIvy
I don't believe there is an "official" definition of "map". It tends to be a 'term of convenience' and typically means a function from one set to another. Depending on the situation, that function might be required to be injective, or surjective, or continuous, ...
• July 5th 2010, 12:27 PM
rainer
Quote:

Originally Posted by HallsofIvy
I don't believe there is an "official" definition of "map". It tends to be a 'term of convenience' and typically means a function from one set to another. Depending on the situation, that function might be required to be injective, or surjective, or continuous, ...

Very interesting.

What about the map for the basic rectangular hyperbola function y=r^2/x. Would that map be surjective, injective, or bijective? I've looked at the definitions for these terms on wikipedia and am guessing bijective, but not sure.

Or, if the following question is easier to anwser then I'd appreciate it equally:

Do the "indifference maps" of microeconomic theory tend to fall under one of these categories generally?

Thanks
• July 5th 2010, 03:11 PM
HallsofIvy
"surjective", "injective", and "bijective" on what sets? It is not even a function "from R to R" because it is not defined for x= 0. And certainly there is no real x such that y= 0. If you mean "from R\{0} to R\{0}" then, yes, it is both injective and surjective so it is bijective. Now, do you know how to prove it is injective and surjective?
• July 6th 2010, 09:09 PM
rainer
Quote:

Originally Posted by HallsofIvy
"surjective", "injective", and "bijective" on what sets? It is not even a function "from R to R" because it is not defined for x= 0. And certainly there is no real x such that y= 0. If you mean "from R\{0} to R\{0}" then, yes, it is both injective and surjective so it is bijective. Now, do you know how to prove it is injective and surjective?

bijective on all sets $\frac{r^2}{x}$ where r>0. I guess you write that...

$R^+\to R^+:x \to \frac{r^2}{x}$

As for proving... Isn't it easier just to prove that it is bijective?

"A function f is bijective if and only if its inverse relation $f^{-1}$ is a function" -wikipedia

Since r is a constant for each function f, there must be a one-to-one correspondence between the set of possible x values and the set of possible $\frac{r^2}{x}$ values. For each x input there is a single f(x) output. This means that the relation between the two sets is functional. And since the set of $\frac{r^2}{x}$ values has no other source but the x values, the relation is total. A relation that is both total and functional is defined as a function.

QED the function is bijective.

(I know there's a fancier way of stating all this but that's basically it, right?)