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Check all the statements that are true:
A. If f:A→Bf:A→B is an injective function and A is finite, then B is finite as well and the cardinality of B is at most the cardinality of A.
B. If f:A→Bf:A→B is a surjective function and B is finite, then A is finite as well and the cardinality of A is at least the cardinality of B.
C. If f:A→Bf:A→B is a surjective function and B is finite, then A is finite as well and the cardinality of A is at most the cardinality of B.
D. If f:A→Bf:A→B is an injective function and A is finite, then B is finite as well and the cardinality of B is at least the cardinality of A.
E. None of the above

Sagot :

Answer: Choice E

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Explanation:

Each item A through D is false because one set may be infinite while the other is finite, or vice versa. One set being finite doesn't automatically make the other finite as well.

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Here's an example of why choice B is false.

A = set of nonzero real numbers

B = {-1, 1}

f(x) = |x|/x

This function is surjective because we target everything in the range B = {-1,1}. Positive x values map to 1, negative x values map to -1. Notice how set A is infinitely large, and B is finite.

The other answer choices can be ruled out through similar logic.

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