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The following reasoning seems correct to me, but is it valid for all sets ( finite and infinite)?

(1) If the range is strictly bigger than the domain,then the domain is trictly smaller that the range.

(2) If the domain is smaller, some element of the domain will have to be related to more than one element of the range in order every element of the range to have a pre-image ( unless it would not belong to the range).

(3) But in that case, it is no longer a function.

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    it definitely matters what you mean by bigger. Cardinality? Lebesgue Measure? Are the 2 most common – NazimJ Nov 13 '19 at 21:33
  • NazimJ. - I mean cardinality. –  Nov 13 '19 at 21:33
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    certainly the range cannot be bigger, in cardinality, than the domain – Masacroso Nov 13 '19 at 21:35
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    It is true and can be proven using axiom of choice (AC). For every point $y$ in the range, choose (via AC) one point $x$ in the domain that maps into $y$. This gives you an injection of the range into the domain, which proves the range is of smaller (or equal) cardinality. I don't know about the status of this claim if you don't assume AC. (I believe, without AC, you cannot even prove that for two sets one is bigger or equal than the other: see https://en.wikipedia.org/wiki/Cardinality#Definition_2:|A|%E2%89%A4_|B| ) –  Nov 13 '19 at 21:36
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    And it looks like it is not necessarily true without enough Choice: https://math.stackexchange.com/a/1055508/221811 – Chappers Nov 13 '19 at 22:16

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