# Number of elements is odd infinity [closed]

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

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

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• I don't understand what "odd infinity number" means – Brenton Nov 17 '15 at 18:36
• 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 – Hrca12 Nov 17 '15 at 18:38
• The term "odd" number applies only to finite numbers. You cannot use it with "infinity". – user247327 Nov 17 '15 at 18:40
• @Hrca12 I suggest looking at this to understand why your question is not well-posed: math.stackexchange.com/questions/49034/… – Brenton Nov 17 '15 at 18:41
• 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. – GEdgar Nov 17 '15 at 18:49

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.

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

• 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. – Asaf Karagila Nov 17 '15 at 18:44
• @Asaf: Interesting - how would the axiom of choice imply every infinite cardinality being even? – Deusovi Nov 17 '15 at 18:47
• Every infinite cardinal satisfies $\kappa=\kappa+\kappa$. – Asaf Karagila Nov 17 '15 at 18:50
• @Asaf Not only that, they are also perfect squares. – Tobias Kildetoft Nov 17 '15 at 18:51
• @AsafKaragila doesn't it mean that infinite cardinals are odd too, since $\kappa + 1 = \kappa$? – lisyarus Nov 17 '15 at 18:54