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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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 $x\ne y.$


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 $B$.

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

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