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Two sets $A$ and $B$ are said to be equivalent, if there exists a bijection between $A$ and $B$.

To prove that $(a,b)$ is equivalent to $(c,d)$

Now I need a map which maps $a$ to $c$ and $b$ to $d$ and all numbers in between. I am thinking that decimal representation of number swill play some role here, like $a.a_1a_2a_3...$ is being mapped to $c.a_1a_2a_3...$ but I am not able to find explicitly such function.

Thanks

Martin Sleziak
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J. Deff
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  • You can simply use a linear function with $a\mapsto c$ and $b\mapsto d$ – Janik Aug 28 '16 at 11:11
  • Note that the mapping $x\to \alpha x$ scales the interval $[a,b]$ by a factor of $\alpha$. Moreover, $x\to x+\beta$ sends the interval on an interval of the same length $\beta$ units further. So it's possible to find a mapping $x\to \alpha x +\beta$ such that $[a,b]$ is sent on $[c,d]$ – H. Potter Aug 28 '16 at 11:12
  • This may be my own preference and 'equivalent' may be the terminology your text, professor or course is using but I like to be very clear in my language in which case I would opt for saying "two sets have the same cardinality or are equinumerous/equipotent/equipollent if there exists a bijection between them". Many branches of math study different equivalence relations between sets determined by the existence of some type of morphism between them so I try to always be clear what exactly the "equivalent" relation I am interested in is. – Prince M Aug 28 '16 at 11:21
  • Try with this: draw in the cartesian plane the two points of coordinates $(a,c)$ and $(b,d)$. Now with a ruler draw the segment connecting these two points. You have just drawn the graph of your bijection! – Crostul Aug 28 '16 at 11:26
  • @Janik Can you elaborate more? – J. Deff Aug 28 '16 at 11:53
  • @H.Potter But without knowing anything about intervals how can we find a mapping or we have to make different cases – J. Deff Aug 28 '16 at 11:55
  • @Crostul How on earth is this? – J. Deff Aug 28 '16 at 12:09
  • See https://en.wikipedia.org/wiki/Linear_interpolation – Crostul Aug 28 '16 at 12:16
  • @Crostul If i apply Lagrange interpolation with these two points i get interpolating polynomial. But i need function – J. Deff Aug 28 '16 at 12:18
  • Isn't the interpolating polynomial a function? – Crostul Aug 28 '16 at 12:25
  • @Crostul Thanks for help. i got it now. btw why downvotes? – J. Deff Aug 28 '16 at 12:27
  • I'm sorry for the downvotes. But it wasn't me. – Crostul Aug 28 '16 at 12:29

1 Answers1

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If $d-c = b-a$ then instantly there's a bijection: just translate one set onto the other.

Otherwise, translate both sets so $a=c=0$. Then multiply everything in the "smaller" set by something appropriate to make sure the two sets' least upper bounds are equal.

Alternative method: we instead show that every bounded open interval bijects with $\mathbb{R}$. This is covered by https://math.stackexchange.com/a/490068/259262 .

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Patrick Stevens
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