To be or not to be

Algebra Level pending

If f : S → T f: S \to T , S = { 1 , 2 , 3 , 4 } S = \{1, 2, 3, 4\} , T = { 5 , 6 , 7 } T = \{5, 6, 7\} , f ( 1 ) = 5 , f ( 2 ) = 6 , f ( 3 ) = 7 f(1) = 5, f(2) = 6, f(3) = 7 , then f f is injective.

False True

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Hobart Pao by Hobart Pao , 23, USA

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1 solution

Hobart Pao
Sep 23, 2016

f f isn't a function. By the definition of a conditional, this is a vacuously true statement.

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My take is that f f is not a function on S S at all, since it fails to map the element 4 4 onto anything.

Brian Charlesworth - 4 years, 8 months ago

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Ah. You're right. I've learnt something new today!

Hobart Pao - 4 years, 8 months ago

(For some reason I can't edit my original comment, so I'll continue this way.)

We could also observe that any legitimate function from one set to another set of a lesser cardinality is necessarily non-injective, (although it can potentially be surjective).

Brian Charlesworth - 4 years, 8 months ago

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