Math Proof/Logic Injection/Surjection Problem

[bronxbombas]

Homework Statement

Determine which of the following statements are true. Give proofs for the true statements and counterexamples for the false statements.

Homework Equations

B. Every non-decreasing function from R to R is injective.

The Attempt at a Solution

So I think I know a function that would suit, but I can't just give a drawing of a graph as a proof. I think it would be a function that would be increasing, eventually go monotone for a little while, and then continue to increase to infinity. Does anybody know a function that would represent this well?

 

[jhicks]

non-decreasing function? Just remember that non-decreasing doesn't necessarily imply increasing. I don't really want to give any more away than that, but a trivial counterexample would work fine; No need to construct some bizarre graph.

 

[bronxbombas]

do you mean something like y=5 would work?

 

[jhicks]

Exactly. It's non-decreasing and it is not injective.

 

[bronxbombas]

I got some more. I'd appreciate some feedback.

C. Every injective function from R to R is monotone.

I think this one's true.

D. Every surjective function from R to R is unbounded.

Would arctan(x) disprove this one? I'm not sure if surjection is limited to the range of a function or for all reals.

E. Every unbounded function from R to R is surjective.

I think this one's true.

 

[jhicks]

C (Dick explains, I failed at the definition of monotone)

Your D also does not work because for every y in R there is not an x in R such that arctan(x) = y. For example, for y = 2 in the codomain, there is no corresponding x.

E I think is the trickiest to construct, but I believe you can make a counterexample by manipulating an obvious unbounded surjection like f(x)=x. See if you can't make that function "skip" a number in the range.

 

[Dick]

C) Suppose the function is discontinuous. D) arctan(x) isn't surjective. E) I think you are again thinking of only continuous functions, and not even all of those.

 

[Dick]

C seems easy to show by contradiction.

Your D also does not work because for every y in R there is not an x in R such that arctan(x) = y. For example, for y = 2 in the codomain, there is no corresponding x.

E I think is the trickiest to construct, but I believe you can make a counterexample by manipulating an obvious unbounded surjection like f(x)=x. How could you disprove this? Try to manipulate f(x)=x so that it doesn't have a f(x)=2. How?

For E) how about f(x)=x^2? C) isn't easy to show because it's not true, they didn't say the function is continuous. D) is on the mark, because there is no counterexample. Surjection means the range is all reals.

 

[bronxbombas]

I understand the example you give for E.

For C, I don't get how that can be false. How can the function be discontinuous if it's from R to R?

I understand why arctan doesn't work for D. So is it safe to assume that one's true?

 

[Dick]

A surjection from R->R covers all reals, it's pretty safe to assume it's unbounded. For C) let f(x)=x for x<0. f(x)=1-x for 0<=x<=1. f(x)=x for x>1. It's injective but it's not monotone. It's also not continuous. It would have to be.