we have f: X→Y is a 1-1 function and Y is countable. Since f is 1-1, is it correct that this implies X ~ f(X) [which is f:X →f(X)] which further implies bijecton? Also, does this imply f(X) ⊆ Y?
we have f: X→Y is a 1-1 function and Y is countable. Since f is 1-1, is it correct that this implies X ~ f(X) [which is f:X →f(X)] which further implies bijecton? Also, does this imply f(X) ⊆ Y?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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we have f: X→Y is a 1-1 function and Y is countable.
Since f is 1-1, is it correct that this implies X ~ f(X) [which is f:X →f(X)] which further implies bijecton? Also, does this imply f(X) ⊆ Y?
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