# Confusion between function and multivalued function.

"What is a function?" can be answered as "Single-valued relations are called functions". But how can "What are the multi-valued function?" be answered?

Will someone clarify my doubt why multi-valued functions are not violating the classical definition of function?

EDIT

This is what Wikipedia says on multivalued functions:

In mathematics, a multivalued function from a domain X to a codomain Y is a heterogeneous relation. However, in some contexts such as the complex plane (X = Y = ℂ), authors prefer to mimic function theory as they extend concepts of the ordinary (single-valued) functions.

In this context, an ordinary function is often called a single-valued function to avoid confusion.

The term multivalued function originated in complex analysis, from analytic continuation. It often occurs that one knows the value of a complex analytic function $$f(z)$$ in some neighbourhood of a point $$z=a$$. This is the case for functions defined by the implicit function theorem or by a Taylor series around $$z=a$$. In such a situation, one may extend the domain of the single-valued function $$f(z)$$ along curves in the complex plane starting at $$a$$. In doing so, one finds that the value of the extended function at a point $$z=b$$ depends on the chosen curve from $$a$$ to $$b$$; since none of the new values is more natural than the others, all of them are incorporated into a multivalued function. For example, let $$f(z)=\sqrt {z}$$, be the usual square root function on positive real numbers. One may extend its domain to a neighbourhood of $$z=1$$ in the complex plane, and then further along curves starting at $$z=1$$, so that the values along a given curve vary continuously from $$\sqrt {1}=1$$. Extending to negative real numbers, one gets two opposite values of the square root such as $$\sqrt {-1}=\pm i$$, depending on whether the domain has been extended through the upper or the lower half of the complex plane. This phenomenon is very frequent, occurring for $$n$$th roots, logarithms and inverse trigonometric functions.

To define a single-valued function from a complex multivalued function, one may distinguish one of the multiple values as the principal value, producing a single-valued function on the whole plane which is discontinuous along certain boundary curves. Alternatively, dealing with the multivalued function allows having something that is everywhere continuous, at the cost of possible value changes when one follows a closed path (monodromy). These problems are resolved in the theory of Riemann surfaces: to consider a multivalued function $${\displaystyle f(z)}$$ as an ordinary function without discarding any values, one multiplies the domain into a many-layered covering space, a manifold which is the Riemann surface associated to $$f(z)$$.

• They are violating the definition of function. Commented Aug 10, 2018 at 18:07

## 3 Answers

Multi-valued functions are violating the (classical) definition of functions.

Multi-valued functions are violating the classical definition of function.

In addition, I can show you an argument why function cannot have more than one different values for each member in the domain of a function.

That is also why the classical definition of function must require no two different values for the same element in the domain of that function.

Suppose

a=f(x) ..................(1)

b=f(x) ..................(2)

~(a=b) .................(3)

From symertircal property of '=',

If b=f(x) then f(x)=b ....(4)

From transitive property of '=',

If a=f(x) and f(x)=b, then a=b....(5)

By modes ponens, (2), (4),

f(x)=b ...................(6)

By modes ponens, (1), (6), and also (5),

a=b (This conclusion contradicts (3), RAA)

Therefore, for two members being as the values of a function, they must be equal to each other.

This derivation is cited from Rudolf Canrap's Introduction to Symbolic Logic and its Application, p. 74

• The first sentence of this answer duplicates the previous accepted correct answer. The second sentence just restates the definition of a function, does not need a proof, and can't be proved since it is just a definition. Please don't answer old questions with accepted answers unless you have something new and correct to add. Commented Sep 13, 2023 at 16:15
• Thanks for your comment. However, I was not intended to 'proof' the definition of a function. I would like to show that why function is defined with this condition so that logical flaws would not be incurred. Commented Sep 13, 2023 at 16:22
• You are not showing anything. All this argument does is invoke the meaning of equality. It has nothing to do with the definition of a function. The "that is why ..." in your post is wrong. Definitions don't require a "why". Commented Sep 13, 2023 at 16:29
• The above cited derivation shown that by assuming a function having two non-identical values for the same argument, one who accept the properties of equality and modus ponens as a rule of inference, having the first three premises, one can deduce that a function cannot have two non-identical values for the same argument. For '''Definitions don't require a "why". ''', do you mean that definitions can be arbitrarily introduced to a mathematical system without considering its consistency? May I ask for a further clarification from your point of view? Commented Sep 13, 2023 at 16:51
• In everyday mathematics the properties of equality and the validity of modus ponens are taken for granted and don't need to be explicitly invoked. You worry about those things only when you are studying mathematical logic for its own sake. The usual definition of a function that relies on the meaning of equality does not need to be checked for consistency. If your argument in this answer were really necessary you would have to add it to just about every answer on this site. Commented Sep 13, 2023 at 18:11

efficiently you can say "relation between variable (or 's') are called functions"

• :it is not correct as single element of a domain can be related more than one element, in that case, your opinion does'nt seem to work Commented Aug 10, 2018 at 19:07