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With more specific saying,

if there is a function that is in continuum from a to b, a and b are real numbers, and bijective to its range, is the number of the function's range infinite?

If it is, are all continuums infinite?

user224661
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  • Welcome to the Math Stack Exchange. Do you mean "are there an infinite number of numbers in the range"? You know that bijective implies that the function is one-to-one and that there are an infinite number of real numbers between any two distinct real numbers. I assume what you mean by "in continuum" is that the function is continuous? – Paul Sundheim Mar 18 '15 at 17:12
  • @PaulSundheim Yes, do all continuous functions have infinite members in range? – user224661 Mar 18 '15 at 17:18
  • No: a constant function is continuous and has finite range (even without mentioning that continuous functions can be defined between any two topological spaces). – A.P. Mar 18 '15 at 17:30
  • Also, what do you mean by "continuum"? And by "infinite" do you mean "has infinitely many elements" or "isn't bounded"? – A.P. Mar 18 '15 at 17:32
  • If the only condition on the function is that it is continuous then the answer is no, as A.P. points out. If the function is required to be both continuous and a bijection then the answer is yes, there are an infinite number of numbers in the range. – Paul Sundheim Mar 18 '15 at 19:08

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You have to be careful... is your function continuous at $x=a$ and $x=b$?

If this function only has to be continuous on $(a,b)$ then, yes, you can find continuous 'infinite functions'.

For example,

$$f(x)=\tan\left(\frac{\pi}{b-a}\left(x-\frac{a+b}{2}\right)\right)$$

is a bijection from $(a,b)$ to $\mathbb{R}$.

If you are continuous on the closed interval $[a,b]$ then the answer is no because the image of a compact set under a continuous function is compact. In other words the best you can do is map $[a,b]$ to another interval $[a',b']\subset\mathbb{R}$.

You can construct a sequence, $\{f_N\}_{N\geq1}$, of continuous and bijective functions from $[a,b]$ to $[-N,N]$ but this sequence does not converge to a continuous function.

JP McCarthy
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