MHB A Surjective function from [0,1]\{1/2} to [0,1]

AI Thread Summary
The discussion revolves around the challenge of finding a surjective function from the interval [0,1] excluding 1/2 to the interval [0,1], with the condition that if f(a) > f(b), then a > b. The original poster expresses difficulty in solving this problem without using cardinality arguments. A participant suggests considering an increasing sequence approaching 1/2 and questions the behavior of the function at that limit. The conversation highlights the complexities of defining such a function under the given constraints.
Ella1
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Hello guys! I'm taking Discrete Mathematics this semester and I got this question in one of my homework tasks.
I've tried thinking about the solution over and over but can't seem to come up with anything..
The question goes like this: Is there a Surjective function from [0,1]\{1/2} to [0,1] such that if f(a)>f(b) that means that a>b?
I must mention that sadly I cannot use any arguments involving cardinality..
Any clue that might help will save my life!p.s Sorry for my poor English :)!
 
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Ella said:
Hello guys! I'm taking Discrete Mathematics this semester and I got this question in one of my homework tasks.
I've tried thinking about the solution over and over but can't seem to come up with anything..
The question goes like this: Is there a Surjective function from [0,1]\{1/2} to [0,1] such that if f(a)>f(b) that means that a>b?
Hi Ella, and welcome to MHB!

Suppose that $(x_n)$ is an increasing sequence in $[0,1/2)$, with $\lim_{n\to\infty}x_n = 1/2.$ What can you say about the point $\lim_{n\to\infty}f(x_n)$? Can it be in the range of $f$?
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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