Is there a real function that is discontinuous on irrational points and continuous on rational point

This is a classic question, and the conclusion is impossible. I just thought about this question, so I accumulated some information :-) to share with LZ

Examples 11 and 13 of the "Borel Sets" section in the first chapter of the book together support this conclusion. The former says that the set of continuous points of the function on the open set is a gδ set, and the latter says that the set of rational numbers is not a gδ set (in fact, the countable set is not a gδ set), and the two can be combined. In addition, it can be proved by using the baire theorem.

There are also elementary methods (of course, real number continuity is necessary), but the process is longer. For example, this problem can be proved with the idea of the Baire theorem. Let's put a reference for LZ Example 2 is the problem of LZ, where a more elementary proof is given (pages 3 to 4).

If it is a general function, it is difficult to get, because the general function is continuous, but the special piecewise function is still possible, as long as the definition field is reasonable.

For example, f(x)=1 (x is a rational number) and =0 (x is an irrational number) is a very special example, which is continuous in the field of rational numbers but not in the field of real numbers. It's a bit strange to say that there is discontinuity in the field of irrational numbers, but I understand it in the field of real numbers.

does not exist. Because according to the definition of continuity, in the process of x approaching a rational point, if it approaches that point in a sequence of irrational points, there is no limit.

Rational numbers, the most common examples are, integers and finite decimals. An integer is a number that can be counted with your fingers. 2 3 4 5 1234 456 12341234 88888888 This finite decimal is the number that can be written entirely after the decimal point.

For example.

All rational numbers can be represented as a natural number.

and the ratio of another natural number, i.e., q=p q

Can rational numbers and rational numbers be listed in their entirety? OK. For example, between and two rational numbers, there must be another rational number and so on.

Irrational number. An example is, pi.

pi=You may forget about it later.

Rational numbers are integers (positive integers.

0, negative integer) and fractions, is the set of integers and fractions in the repentance.

An integer can also be seen as a denominator.

is a fraction of one. Real numbers that are not rational numbers are called irrational numbers, i.e., the fractional part of an irrational number is an infinite number that is not cyclical. It is one of the important contents in the field of "number and algebra", which has a wide range of applications in real life, and is to continue to learn real numbers and algebraic formulas.

Fangpei shed, inequality, Cartesian coordinate system.

Functions, statistics and other mathematical disciplines and the foundation of related subject knowledge.

The set of rational numbers can be represented by the uppercase black orthography symbol q. But q does not denote a rational number, and a set of rational numbers and a rational number are two different concepts. A rational number set is a set of elements that are all rational numbers, while a rational number is a set of all the elements in a rational number set.

Hehe: It's not easy.

Proof 1: Rational numbers are countable, for example, p q is equivalent to the set of natural numbers, so the numbers are equal; are all countable, i.e., discretely distributed on the number axis, so there are discrete discontinuities for rational numbers.

And the point of the front answer of the upper hall on the base closed number axis is continuous, so r-q is continuous.

Rational numbers are ordered, but irrational numbers are disordered. In other words, what is absolute is always finite, and what is not absolute is always infinite.

How do you define continuous in a set of nuclei in numbers? If the lack of front digging is like the first floor said, then at most it can only be said that there are rational numbers to count, so it is the so-called "order", and the specific number method is the cantor diagonal method. Irrational numbers cannot be counted, and there are many ways to use the counterargument method, such as the binary construction method.

However, the unbasic number does not mean that the irrational number is out of order. The general set of numbers and the less than relation constitute a fully ordered set, how can it be said to be disordered? By the way, rational numbers and irrational numbers are dense and infinite.

This statement should be made only in probability theory.

was established. It can be proved that the rational point is on the number line.

has a chance of zero, and a chance of taking any point as an irrational point is 1

But there are dozens of branches of mathematics, so the theory of inference is not true in other branches.

First of all, the discontinuity of the rational point is provable, but is the irrational point continuous?

Set theory. Cantor, the founder of the company, famously proposed the famous Contor conjecture: there is no other potential between two successive potentials.

In 1963, the American mathematician Cohen proved that this conjecture is independent of the set theory system. That is, this conjecture can never be proven.

So irrational numbers.

The question of continuity is like the fifth axiom of geometry and the axiom of choice in set theory--- which can neither be proved nor refuted.

In other words, in set theory, defining the continuity of irrational numbers does not lead to contradictions, and defining discontinuity still does not lead to contradictions. Therefore, it shows that the results proved in probability theory do not hold true in set theory.

Because someone asked in the discussion of Dongqiao, I would like to make a supplementary explanation.

First of all, explain what "many" is. Rational and irrational numbers are not equal, that is, a one-to-one correspondence cannot be established. If two sets can establish a one-to-one correspondence, they are said to be equal (i.e., "as many").

The equivalence of an infinite set may be intuitively different as the equivalence of a finite set, e.g., integers and even numbers can correspond one-to-one (n corresponds to 2n), so they are equivalent.

Since it is a rational number that can be written as an integer fraction, rational numbers and integer pairs are equivalent; And because integer pairs (0,0), (0,1), (1,0), (1,1) ......It can be arranged in an ordered column (positive and negative can be staggered), so integer pairs and natural numbers are also equal.

Similarly, since irrational numbers have ,,, a part of an irrational number can establish a one-to-one correspondence with a natural number, they are equivalent. Thus irrational numbers are no less than natural numbers, and therefore no less than rational numbers.

All we need to do now is show that irrational numbers are not equivalent to natural numbers.

We use counter-evidence. Inverted irrational numbers can be arranged in a column (and thus numbered ......

We can find a new irrational number whose first digit is different from the first number in the sequence above, and the second digit is different from the second number in the sequence,......Thus this new irrational number is not in the sequence, which is a contradiction. This contradiction shows that irrational numbers cannot be arranged in a single column, i.e., there are more irrational numbers than natural numbers, and thus more than rational numbers.

What do you want me to say......

How do you think irrational numbers compare with Kaili numbers? There is no accurate numerical guess for irrational numbers, no matter which interval is infinite, and there are also countless rational and respectful numbers.

You're a genius if you can compare!!

Let q: n q q(n)=r ,r q.

Let g: q q g(q(n))=2*-n.

Let f: r q f(x) = g(r), r x , r q.

Then the function f defined above is a function defined on the set of real numbers, which is interrupted at the point of rational numbers and continuous at the point of irrational numbers.

Let a be a rational point, then for any >0, there is δ>0, when |x-a|<δ, |f(x)-f(a)|To meet |x-a|<δ irrational point b, when x satisfies |x-a|<δ, |f(x)-f(b)|=|f(x)-f(a+f(a)-f(b)|≤/2|f(x)-f(a)|+f(b)-f(a)|So f(x) must be continuous at the irrational point b.

Therefore, there is no such thing as a discontinuous function at all irrational points.

For example, a is the maximum number and b is the minimum number.

That option A' =a+1 >a

and b' = b-1 < b

So A has no maximum number, and B has no minimum number.