Show Explicit Bijection Between Sets (0,1)

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To show an explicit bijection between sets, one must define a function that pairs each element of one set uniquely with an element of another set. For example, a bijection can be illustrated with finite sets, such as mapping A = {0, 1, 2, 3, 4} to B = {8, 9, 10, 11, 12} using the function f(a) = a + 8. When discussing the interval (0, 1), it is essential to specify the target set for the bijection, as a bijection must connect two distinct sets. The conversation emphasizes the need for clarity in defining both sets involved in the bijection. Understanding the concept of bijection is crucial for establishing these mappings effectively.
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How can I show an explicit bijection between sets?
anyidea?
how about with (0,1)?
 
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You just write down a function. Do you know what the definition bijection is?

For example, if you want to show a bijection between A = {0, 1, 2, 3, 4} and B = {8, 9, 10, 11, 12} you could write
f: A \to B, f(a) = a + 8
(for finite sets, you can also write it out as: define f by f(0) = 8, f(1) = 9, ..., f(4) = 12).

how about with (0,1)?
How about it? If you mean: how about a bijection, you are asking half a question. A bijection is always between two sets. "How about a bijection from (0, 1) to (3, 4)", or "from (0, 1) to (0, 1)", or "between (0, 1) and (-2, 2)" would make sense.
 
Compuchip is correct. Think over it.
 

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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