Answered

Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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

====================================================

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.

---------

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.

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.