2
$\begingroup$

Let f be an injective function, that is:

$f : X \rightarrow Y$

$f(a) = f(b) \implies a = b$

Now, my question is, does the following need to hold in order for function to be injective:

$(\forall x \in X)(\exists y \in Y) (x,y) \in f$ (if we consider function to be a set of ordered pairs) (denote this statement by $(*)$)

1st case

If it does, then in order for function to be bijective, it needs also to be injective. For example in this case the function $f(x) = \frac 1 x$ is not injective, because it is not defined for x = 0 and $(*)$ does not hold; therefore it's also not bijective and inverse function does not exist. But $f^{-1}(x) = \frac 1 x$ .

2nd case

If statement $(*)$ is not required for function to be injective, then the definition of bijective function to be one-to-one correspondence does not hold, since there can be elements in domain that are not paired with any elements in range. So if we wanted to keep the definition of injective function without $(*)$, we would than have to redefine or rather extend bijective function not only as injective and surjective, but also satisfying $(*)$.

It seems to me to be intuitive paradox, but I'm sure I have made a mistake somewhere and I'd be greatful if someone explained it to me :D

$\endgroup$
1
  • 1
    $\begingroup$ That condition needs to hold in order for $f$ to be a function. $\endgroup$ Aug 10 '12 at 20:49
1
$\begingroup$

If for a given $x\in X$ there is no $y\in Y$ s.t. $(x,y) \in f$ then $x$ isn't in the domain of $f$. At that point saying $f:X\rightarrow Y$ is somewhat misleading. The domain of $f(x)=\frac{1}{x}$ is $\mathbb R\backslash \{0\}$.

$\endgroup$
2
  • $\begingroup$ Oh so in the very definition of $f: X \rightarrow Y$ it has to be that it is defined $\forall x \in X$... $\endgroup$ Aug 10 '12 at 14:56
  • $\begingroup$ Yes. In general, it doesn't make sense to talk about the properties of a function outside of its domain. For instance, we can't really ask if $f(x) = \frac{1}{x}$ is continuous at 0. $\endgroup$
    – axblount
    Aug 10 '12 at 15:27
1
$\begingroup$

The definition of a function is $$ \forall x \in X\ \exists! y \in Y:\ (x,y) \in f. $$ The problem in your claim is that a map $$ \mathbb{R} \rightarrow \mathbb{R}, x \mapsto x^{-1} $$ does not exist. Thats why you have to specify $X$ and $Y$.

$\endgroup$
5
  • 1
    $\begingroup$ Ok, that pretty much solvs my problem, since I was Taught that your statement no1 does not have to hold :) thx $\endgroup$ Aug 10 '12 at 14:57
  • $\begingroup$ For a map/function it has to hold. In school they often teach that you have to specify the domain by the function, but that's bad practise. Exercises like this can better be given like: Find the maximal domain of a mapping as a subset of certain set when the range is this set. I guess that's to complicated for school. $\endgroup$
    – sebigu
    Aug 10 '12 at 15:01
  • $\begingroup$ No I'm doing this out of personal interest not for school but thanks anyway. I probably just skipped the line in the book where it specifies that function has to be defined for every x in the domain, since i always try to find out why would the mathematical object be defined the way it is... and now i see that this is necessary for functions to work :P $\endgroup$ Aug 10 '12 at 15:08
  • 1
    $\begingroup$ For a function, the definition is actually much more restrictive -- the $y$ has to be unique. $\endgroup$
    – gt6989b
    Aug 10 '12 at 19:40
  • $\begingroup$ You are right, I will edit the answer. $\endgroup$
    – sebigu
    Aug 10 '12 at 20:41
0
$\begingroup$

I think any bijection is defined to be both a surgection (onto) and an injection (1-to-1) at the same time.

As for your example, it matters to which sets you are mapping. For example, if we consider $f(x) = 1/x$ as a map $f:\mathbb{R}^+ \to \mathbb{R}^+$, all conditions for a bijection hold. Similarly for $\mathbb{R}-\{0\}$. However, for $\mathbb{R}$, the relation $f$ is not just a bijection, but it is not even a function, since functions require each input to be related to exactly one output, and $f(0) \not \in \mathbb{R}$.

One way to fix the problem is to consider $\mathfrak{R} = \mathbb{R} \cup \{\pm \infty\}$ and then $f$ is an injection (but not a surjection and not a bijection) on $\mathfrak{R}$. Finally, if you want a bijection, you may want to consider $f$ on $\mathfrak{R} = \mathbb{R} \cup \{\infty\}$ (or on $\mathbb{R}^+$ or $\mathbb{R}^-$ as we have already pointed out).

$\endgroup$
1
  • $\begingroup$ thx, i had a flaw in basic definition of a function's domain :) $\endgroup$ Aug 10 '12 at 15:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.