Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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.

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.