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?
Chapter3: Functions
Section3.3: Rates Of Change And Behavior Of Graphs
Problem 2SE: If a functionfis increasing on (a,b) and decreasing on (b,c) , then what can be said about the local...
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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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