3
$\begingroup$

This is a different but related question to one I asked earlier. I link to it here:

"To show that f is injective" - I don't get this statement

I am pretty new to "functions" having just went through a quick primer on "propositional logic". So the $\rightarrow$ symbol which represents a conditional statement looks very similar in the definition of injective below.

Suppose that $f: A \to B$ (Is this to be read in English as "If $A$ is true then $B$ is true"?)

To show that $f$ is injective - Show that if $f(x) = f(y)$ for arbitrary $x, y \in A$ with $x \neq y$, then $x = y$.

How do I read this in English, specifically the part where there is a comma. I am not sure if this is stating that the ordered pair $x,y$ is an element of set $A$ or just the $y$ element itself. If the author of the text (Rosen) is talking about $x, y$ as an ordered pair then it would help to use parentheses.

$\endgroup$
  • 2
    $\begingroup$ The arrow in "$f:A \to B$" is different from the arrow in "$P \implies Q$". The first arrow means that $f$ is a function from $A$ to $B$; the second arrow means that $P$ implies $Q$ (or equivalently: if $P$ is true then $Q$ is true). $\endgroup$ – TonyK Sep 20 '14 at 15:59
  • 1
    $\begingroup$ It's worth noting that propositional logic formulas frequently use $\to$ rather than $\implies$ for the implication operator, as opposed to entailment: en.wikipedia.org/wiki/Material_conditional $\endgroup$ – Erick Wong Sep 20 '14 at 16:49
2
$\begingroup$

The first part should be read "Suppose that $f$ is a function from the set $A$ into the set $B.$"

The second part doesn't make sense, as written. It should say "Show that if $f(x)=f(y)$ for arbitrary $x,y\in A,$ then $x=y.$" That is, if $f(x)=f(y)$ implies that $x=y$ for any elements $x$ and $y$ of $A$ (not necessarily different elements, just with different names), then $f$ is injective. The idea, here, is that a given function output will only be the result of a single function input. Put another way, if $f$ is injective, then we will never have a situation where $f(x)=f(y)$ and $x\ne y.$

$\endgroup$
2
$\begingroup$

It reads as: "Show that if $f$ of $x$ equals $f$ of $y$ for arbitrary $x$ and $y$ in $A$ with $x$ different than $y$, then $x$ equals $y$". The comma as nothing to do with pairs, it is just a comma from common language.

Moreover $f \colon A \to B$ should be read as: $f$ is a function from the set $A$ into the set $B$.

$\endgroup$
  • $\begingroup$ Thank you very much. Very helpful answer. $\endgroup$ – Bradley Sep 20 '14 at 15:58

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.