- #1
alexfloo
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I'm currently reading Sterling Berberian's Foundations of Real Analysis, and the first chapter had an overview of foundational mathematics from axiomatic set theory to constructive proof of the real numbers. I was looking over this chapter, and I found this exercise in the functions section:
- If [itex]f\circ h=g\circ h[/itex] and h is surjective, then f=g.
He notes afterwards that this cancellation law would fail if we admit the empty function. However, the way I see it empty functions are as follows:
The first two cases, regardless of the initial and final sets, always have the same graph: the empty set. Admitting these (as far as I can tell) don't change any of the properties of functions. The third case, however, even if we "admitted" empty functions still would not exist, because by the definition of a function, every member of the initial set must be mapped to exactly one member of the final set. There are no members of the final set, however. That is, if our definition of the function made no mention of empty function one way or another, and you told me that:
[itex]f:X\rightarrow Y[/itex]
I can logically conclude that:
[itex](Y=\{\})\Rightarrow (X=\{\})[/itex].
Essentially, although empty functions are degenerate, I don't see that the actually complicate the machinery. One of the three types is problematic, and that particular type is tautologically nonexistent.
Am I missing something, or did Berberian likely just choose not to with all this because it's rather trivial anyways?
- If [itex]f\circ h=g\circ h[/itex] and h is surjective, then f=g.
He notes afterwards that this cancellation law would fail if we admit the empty function. However, the way I see it empty functions are as follows:
- Functions from the empty set to the empty set
- Functions from the empty set to a nonempty set
- NOT functions from a nonempty set to the empty set
The first two cases, regardless of the initial and final sets, always have the same graph: the empty set. Admitting these (as far as I can tell) don't change any of the properties of functions. The third case, however, even if we "admitted" empty functions still would not exist, because by the definition of a function, every member of the initial set must be mapped to exactly one member of the final set. There are no members of the final set, however. That is, if our definition of the function made no mention of empty function one way or another, and you told me that:
[itex]f:X\rightarrow Y[/itex]
I can logically conclude that:
[itex](Y=\{\})\Rightarrow (X=\{\})[/itex].
Essentially, although empty functions are degenerate, I don't see that the actually complicate the machinery. One of the three types is problematic, and that particular type is tautologically nonexistent.
Am I missing something, or did Berberian likely just choose not to with all this because it's rather trivial anyways?