Implication in Injective function?

[Mohammed Mujamil]

"As I understand it, the '⇒' symbol (which stands for the word 'implies') means the following: Given two statements A and B:

( ( if (A == true) then (B == true) )

or

(if (A == false) then ((B == true) or (B == false))) )

or in words: If Statement A is true, B must also be true, though if A is false, B can be either true or false.

The definition of an injective function according to my book and wikipedia is: f is called injective, if the following holds true: f(a)=f(b)⇒a=b Which means if we apply two elements a,b∈A to f:A→B, and the resulting two elements f(a) and f(b)∈B are equal, then the original two values a,b must be the same element, hence a=b.

Now comes the part that confuses me: If we apply the definition above to the other, we get the following:

if f(a)=f(b) then a=b
or

if f(a)≠f(b) then a=b or a≠b
So the last line says if f(a)≠f(b) then a and b can either be the same element or different elements. The latter is fine, if f returns two unique elements, the elements a,b applied to the function must also be unique, hence a≠b. But its also possible for a=b, which would mean that it is possible to get more than one unique element from f, according to the definition.

That last conclusion is of course wrong, since that would be a property of a surjective function and not an injective one. Thanks for your time."

 

[Dr.Peterson]

The part where you say "(if (A == false) then ((B == true) or (B == false))) )" says nothing. It says only that if A is false, we can say nothing about B; anything could happen. So it is not needed in the definition of implication.

In particular, if f(a) is known to equal f(b), then (assuming f is injective) we can conclude that a=b. But if f(a)≠f(b), then a and b might be equal, or not, as far as this implication is concerned.

But since f is a function, if f(a)≠f(b), then a can't equal b. (It has nothing to do with being surjective!)

And this doesn't cause any trouble, because the statement that a=b can be either true or false doesn't mean that both cases can actually happen; it only means that we can't know which, based on the implication under discussion.

 

[pka]

The definition of an injective function according to my book and wikipedia is: f is called injective, if the following holds true: f(a)=f(b)⇒a=b Which means if we apply two elements a,b∈A to f:A→B, and the resulting two elements f(a) and f(b)∈B are equal, then the original two values a,b must be the same element, hence a=b.
Now comes the part that confuses me: If we apply the definition above to the other, we get the following:
if f(a)=f(b) then a=b
or
if f(a)≠f(b) then a=b or a≠b
So the last line says if f(a)≠f(b) then a and b can either be the same element or different elements. The latter is fine, if f returns two unique elements, the elements a,b applied to the function must also be unique, hence a≠b. But its also possible for a=b, which would mean that it is possible to get more than one unique element from f, according to the definition. That last conclusion is of course wrong, since that would be a property of a surjective function and not an injective one. Thanks for your time."

This may be more than you need or want,
If \(A~\&~B\) are sets then the statement that \(f:A\to B\) is a function from \(A\) to \(B\) means:
1) \(f\subseteq A\times B\)
2) if \(a\in A\) then \(a\) is the first term in some pair of \(f\)
3) no two pairs in \(f\) have the same first term.
Moreover:
If \(f\) is a injection means that no two pairs in \(f\) have the same second term.
If \(f\) is a surjection means that each element of \(B\) is the second term of some pair in \(f\).

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