# Why does the definition of a one-to-one function contain an implication rather than an equivalence?

Reading Tao´s Analysis, I have seen the following definition of one-to-one functions:

I am thinking about something rather simple here: Shouldn´t the implication be an equivalence instead? I thought about the following: We have the statement: $$x \neq x´ \rightarrow f(x) \neq f(x´)$$. Then we have the four cases for every $$x$$,$$x´$$ pair:

• $$x \neq x´$$ is true, $$f(x) \neq f(x´)$$ is true. This is always true for every $$x$$,$$x´$$ due to one-to-one property.
• $$x \neq x´$$ is true, $$f(x) \neq f(x´)$$ is false. This is always false for every $$x$$,$$x´$$ due to one-to-one property.
• $$x \neq x´$$ is false, $$f(x) \neq f(x´)$$ is true. This is always false for every $$x$$,$$x´$$ since obviously for every $$x = x´$$ we have $$f(x) = f(x´)$$ due to function definition.
• $$x \neq x´$$ is false, $$f(x) \neq f(x´)$$ is false. This is always true for every $$x$$,$$x´$$ due to the above item.

So in general $$x \neq x´$$ is always true when $$f(x) \neq f(x´)$$ is true and it is false when $$f(x) \neq f(x´)$$ is false. This should lead to an equivalence instead of implication.

What am I missing here, why an implication is being used instead of equivalence?

• The implication $x=x' \Longrightarrow f(x)=f(x')$ is the definition of function, so it's oblviously and omitted, but yes, if $f$ is injective, then it's an equivalence. Commented Aug 10 at 21:23
• Please do not use images to convey critical information not otherwise present in your post Here are the reasons why. Commented Aug 10 at 23:06
• Adding on to @SigmaAlgebra's comment: Definitions are usually given using the fewest/ weakest conditions possible, as that makes it easier to show that those conditions are met. Commented Aug 11 at 1:58

It would be interesting to state the condition as an equivalence, that is to add the other direction of the implication, if it could be the case that $$f(x) \neq f(y)$$ but $$x = y$$. However, if this was the case you would have the same element of the domain, going to two different elements of the co-domain; i.e., you would not have a function anymore. Therefore, by the moment you are talking about functions, it is assumed that such a condition is always satisfied. The condition of injectivity is thus stated as the only thing that could hypothetically fail.

Belabouring Sigma Algebra's and JonathanZ's comments:

A function $$f$$ is one-to-one iff $$f(x)=f(x')\implies x=x'$$

means

• if $$f$$ is a function, then $$f \text{ is }\textbf{one-to-one} \iff \big(f(x)=f(x')\implies x=x'\big),$$

which mathematically (by the definition of function) implies

• if $$f$$ is a function, then $$x=x'\implies f(x)=f(x')$$ and $$f \text{ is }\textbf{one-to-one} \iff \big(f(x)=f(x')\implies x=x'\big),$$

which logically implies

• if $$f$$ is a function, then $$f \text{ is }\textbf{one-to-one} \iff \big(f(x)=f(x')\iff x=x'\big),$$

that is,

• A function $$f$$ is one-to-one iff $$f(x)=f(x')\color\red\iff x=x'.$$

With the definition of a function tacit, the desired equivalence (in red) is considered implicit in the given definition, or at least trivial to derive. And since the missing converse implication is not pertinent in proving that a function is one-to-one (as opposed to proving than an object is a one-to-one function), the crisper given definition is preferable to the last bullet point.

The equivalence of two statements given in the theorem is valid because in logic, we have:

$$(A\implies B)$$ equivalent to: $$\neg B\implies \neg A$$

Which can be proved by

$$(A\implies B)\iff\neg A\vee B$$

Which can be proved by drawing a truth table with these elementary charts:

The same logic charts is the same as saying two logical expressions are equivalent.

And yes, since that one value in the domain (x) can not be associated with multiple values based on the definition of functions, so given $$x_1,x_2\in X, x_1=x_2\implies f(x_1)=f(x_2)$$(given that $$X$$ is the domain)

And yes, given $$f$$ one-to-one (injective), we can have $$x_1=x_2\iff f(x_1)=f(x_2)$$

As for the reason why this is not shown like in @JonathanZ 's comment. It is due to that the by stating $$f$$ is a function, it already implies that $$x_1=x_2\implies f(x_1)=f(x_2)$$, else it wouldn't be a function. And that theorems don't typically repeat things, if they assume you know this thing from earlier on.

Think of a general function $$f(x)$$ as an input-output relationship. By definition of a function, the same input must have the same output.

• Injective (aka one-to-one): An output can only have one input, meaning each output is associated with at most one input. No two different inputs can produce the same output.

• Surjective: All outputs have at least one input, meaning every element in the output set is the image of at least one element from the input set.

• Bijective (aka one-to-one correspondence): Each input has a unique output, and each output is the result of exactly one input. This means the function is both injective and surjective, forming a perfect pairing between the input and output sets.

There are many answers here explaining that the equivalence indeed holds. I would like to go in a different direction: This answer is about mathematical style.

Even though it may seem like it to a beginner student, math is not only about cold, hard logic in the sense that we try to pump out any and all true statements we can find. Just like your favorite book or newspaper don't put every bit of true information directly into the text. Only those bits the author deems valuable for what he wants to achieve.

The same is true in writing math, and by inclusion it's true for writing definitions. The intent of the author is not usually to dump every bit of truth about the defined object into the definition. Instead, there are two possible intentions when writing a definition:

• Make it minimal. You see this when people write stuff like "A non-empty subset $$H\subseteq G$$ of a group $$G$$ is called a subgroup if it is closed under the group operation and taking inverses". There is more to say about subgroups, for instance, they always contain the neutral element of the group. But the definition is given in such a way that there are fewer conditions to verify. Sure, if $$H$$ is a subgroup, then it must contain the neutral element. But I don't have to verify that in order to show that it's a subgroup, because I got a minimal definition.
• Make it clear what the main idea behind the definition is. This also requires trimming some unnecessary fluff, but is sometimes different to the intention of minimalizing stuff. For instance, I dislike the definition of subgroups above, because it doesn't even mention the point of what a subgroup is supposed to be: A subset that also happens to be a group, using all the same group structure. So I would define: If $$(G,\cdot,e)$$ is a group and $$H\subseteq G$$, then $$(H,\cdot\vert_{H\times H},e)$$ is called a subgroup of $$G$$ if it is a group. This is harder to verify, and it would be a good idea to prove that the minimal definition is equivalent because it's easier to work with. But this one is closer to the heart of the matter.

Now how does this pertain to your issue? Well, both possible intentions lead to the definition with just one implication.

• If we want to make the definition minimal, then we would like to include only the weakest possible statements. If the equivalence already follows from the implication, then we write down the implication only, because then we have one less thing to verify if we want to show that some function is one-to-one.
• If we want to get to the heart of the matter, we need to identify what makes a one-to-one function special. The implication $$f(x)\neq f(x')\Rightarrow x\neq x'$$ is not special. It's true for every function. If we want to make it clear what makes one-to-one functions special, we should not emphasize things it has in common with all the other functions, only the things that make it different from all others. And that's just the implication $$x\neq x'\Rightarrow f(x)\neq f(x')$$.