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If we have one continuous function F(x), and if we define f(x)=F(x) on domain from open interval (a, b), and if F(a)=F(b)
If function f(x) is monotonically increasing from point a to point M, and monotonically decreasing from point M to point b
Can we assume there is odd infinity number of numbers for which f(x) is defined?
Because for every x there is one z where x,z are from domain of f(x) except for M?
Every element from that range has it's pair except for M

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closed as unclear what you're asking by Morgan Rodgers, hardmath, user149792, user147263, mrf Nov 17 '15 at 23:10

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    $\begingroup$ I don't understand what "odd infinity number" means $\endgroup$ – Brenton Nov 17 '15 at 18:36
  • $\begingroup$ We know that there is infinitely many numbers from point a to b, but can we say that we are sure that there is odd number of that from all of that given above $\endgroup$ – Hrca12 Nov 17 '15 at 18:38
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    $\begingroup$ The term "odd" number applies only to finite numbers. You cannot use it with "infinity". $\endgroup$ – user247327 Nov 17 '15 at 18:40
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    $\begingroup$ @Hrca12 I suggest looking at this to understand why your question is not well-posed: math.stackexchange.com/questions/49034/… $\endgroup$ – Brenton Nov 17 '15 at 18:41
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    $\begingroup$ An interesting thing that we can prove: a continuous function defined on $(a,b)$ cannot be exactly 2-to-1. But we don't prove it by talking about "even" and "odd" infinity. $\endgroup$ – GEdgar Nov 17 '15 at 18:49
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If your open interval has a cardinality which is odd infinity then presumably a similar argument would suggest that a half-open interval would have a cardinality which is an even infinity.

It is possible to find a bijection between a half-open interval and an open interval, or between a half-open interval and a closed interval: see How to define a bijection between $(0,1)$ and $(0,1]$? and the questions linked form it.

The bijection shows that these two cardinalities are in fact the same, which is why infinities are not described as odd or even.

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Infinity cannot be even or odd. "Odd infinity" is a nonsensical phrase.

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    $\begingroup$ That's not true. Assuming the axiom of choice every infinite cardinality is even; and that's quite odd. More specifically in asking "how many numbers satisfy blah" we essentially talk about cardinality rather than the intentionally "less defined" notion of infinity in calculus. $\endgroup$ – Asaf Karagila Nov 17 '15 at 18:44
  • $\begingroup$ @Asaf: Interesting - how would the axiom of choice imply every infinite cardinality being even? $\endgroup$ – Deusovi Nov 17 '15 at 18:47
  • $\begingroup$ Every infinite cardinal satisfies $\kappa=\kappa+\kappa$. $\endgroup$ – Asaf Karagila Nov 17 '15 at 18:50
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    $\begingroup$ @Asaf Not only that, they are also perfect squares. $\endgroup$ – Tobias Kildetoft Nov 17 '15 at 18:51
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    $\begingroup$ @AsafKaragila doesn't it mean that infinite cardinals are odd too, since $\kappa + 1 = \kappa$? $\endgroup$ – lisyarus Nov 17 '15 at 18:54

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