Very discontinuous function

Relevant wiki: Cardinality

Here is a proof by contradiction that no function such as $f(x)$ exists.

First assume that $f(x)$ exists and define another function $h:\mathbb{R} \rightarrow \mathbb{Z} | \space h(x)=\lfloor f(x) \rfloor$

Since (by the condition of the problem) all values of f(x) differ by at least one, we see that h(x) takes a different integer value for every real number.

The function $h(x)$ thus sets up a 1-1 correspondence between the real numbers and a (possibly improper) subset of the integers. But we know (since Cantor ) that such a correspondence is impossible.

And so our starting assumption - that $f(x)$ exists - must be incorrect.

The function f takes only one value on the intervals [k, k+1) for all integers k, therefore, the set of its values is at most countable.

Suppose that such a function $f(x)$ exists. $f$ must be one-to-one (injective), because if it were not, then there would be two distinct values in its domain that output to the same element in its range, contradicting the condition that $|f(a) - f(b)|\geqslant 1$ . We can therefore assume that $f$ is always increasing, although the same argument will hold if it is always decreasing.

Let's examine two distinct real numbers $m$ and $n > m$ . We see that $f(m)$ differs from $f(n)$ by some constant $C_1$ that is greater than or equal to 1. We also see that $f(m)$ differs from $f\left(\frac{m + n}{2}\right)$ by some constant $C_2$ that is greater than or equal to 1. And we also see that $f\left(\frac{m + n}{2}\right)$ differs from $f\left(m + \frac{n - m}{4}\right)$ by some constant $C_3$ that is greater than or equal to 1. And we also see that $f\left(m+ \frac{n- m}{4}\right)$ differs from $f\left(m + \frac{n-m}{8}\right)$ by some constant $C_4$ that is greater than or equal to 1. In fact, we can continue defining such constants ad infinitum . Therefore, the absolute difference between $f(m)$ and $f(n)$ must be greater than $C_1 + C_2 + C_3 + \cdots \geqslant 1 + 1 + 1 + \cdots.$ However, this series diverges, meaning that the difference in the distinct outputs of the function must be infinitely great. Therefore, reductio ad absurdum, the function cannot exist.

Please let me know if my reasoning is correct:

I consider the case where a and b can be made arbitrarily close, while in keeping with the conditions in the question, f(b) - f(a) ≥ 1. As b approaches a, f(b) - f(a) is still greater than or equals to 1, which means that f(b) - f(a) / (b-a) will go to infinity as b approaches a. Hence, the graph of such a function will be a vertical line, in which case, it is no longer a function.

Therefore, such a function cannot possibly exist.

We know the real numbers are uncountably infinite. By contradiction, assume such an f did exist. Then it should be obvious you can form a function g that takes the range of f to the natural numbers. Therefore the function h=gf takes the real numbers to the natural numbers, thus contradicting our claims of R being an uncountable set. See Cantor's diagnolisation proof to see that R is uncountable, though this should be somewhat intuitive.

By using LMVT

As it is the function of real numbers to real numbers there may be the situation such that both "a" and "b" will be the same number and then it will not satisfy the condition.

Since the set of real numbers is uncountably infinite, no function can map it the merely countable infinite set of integers.

When $a>b$ , let me set $a-b=dx$ and $f(a)-f(b)=dy$ . Then $dy \geq dx+1, \dfrac{dy}{dx} \geq 1+\dfrac{1}{dx}.$

But since $dx$ goes to $0$ , $1+\dfrac{1}{dx}$ goes to infinity, which makes explicit function $f(x)$ impossible.

Assuming this function exists, that means you could map every real number in R Where |R| = א to a group that is countable, as in every real number having a unique whole number, where the power of the whole numbers Z is א0 hence because א > א0 there is no function that maps R to Z injectjvly and fully.