- #1
StatOnTheSide
- 93
- 1
This is in reference to the Chapter 10. of P. R. Halmos's Naive Set Thoery, namely 'Inverses and Composites'. He speaks of two equations in that section.
1. Let f be a function from X to Y. The inverse of a set is defined as another set as follows.
f[itex]^{-1}[/itex](B) = {x[itex]\in[/itex]X:f(x)[itex]\in[/itex]B}.
The first equation is
f[itex]^{-1}[/itex](Y-B) = X-f[itex]^{-1}[/itex](B)
It can be shown that the above is always true.
2. In the excercises, it is said that
Y — f(A)=f(X — A) if and only if f is one-to-one and onto.
By f(A), he means all the elements y[itex]\in[/itex]Y such that f(x)=y for x[itex]\in[/itex]A.
My question is this. Replacing f by f[itex]^{-1}[/itex] and changing X to Y and Y to X gives us equation in 1. above. What is it about f[itex]^{-1}[/itex] that makes 1. true all the time?
I understand that by definition, f[itex]^{-1}[/itex] is onto i.e there is no x in X which does not have an image in Y under f; however, f[itex]^{-1}[/itex] is not really one-to-one.
The main problem I have is with the one-to-one-lessness of f[itex]^{-1}[/itex]. Inspite of f[itex]^{-1}[/itex] not being one-to-one, it is turning out to be true while 2 requires any f to be one-to-one.
Now you may ask me as to why f[itex]^{-1}[/itex] is not one-to-one. Let B1 and B2, subsets of Y, be such that range of X under f is a subset of both B1 and B2. Also, let B1[itex]\neq[/itex]B2. But both of them map to X under f[itex]^{-1}[/itex]. Please note that range of f[itex]^{-1}[/itex] is defined to be a set while f, according to Halmos's convention, can operate on a set OR on elements and hence, can have a range whose elements are sets or elements.
In this context, range of f is a set of sets as it is operating on sets. Strictly speaking, we are looking here at f operating on P(X) to P(Y) while f[itex]^{-1}[/itex] operates on P(Y) to P(X).
Kindly let me know as to why 1 is true inspite of the fact that f[itex]^{-1}[/itex] is not one-to-one.
1. Let f be a function from X to Y. The inverse of a set is defined as another set as follows.
f[itex]^{-1}[/itex](B) = {x[itex]\in[/itex]X:f(x)[itex]\in[/itex]B}.
The first equation is
f[itex]^{-1}[/itex](Y-B) = X-f[itex]^{-1}[/itex](B)
It can be shown that the above is always true.
2. In the excercises, it is said that
Y — f(A)=f(X — A) if and only if f is one-to-one and onto.
By f(A), he means all the elements y[itex]\in[/itex]Y such that f(x)=y for x[itex]\in[/itex]A.
My question is this. Replacing f by f[itex]^{-1}[/itex] and changing X to Y and Y to X gives us equation in 1. above. What is it about f[itex]^{-1}[/itex] that makes 1. true all the time?
I understand that by definition, f[itex]^{-1}[/itex] is onto i.e there is no x in X which does not have an image in Y under f; however, f[itex]^{-1}[/itex] is not really one-to-one.
The main problem I have is with the one-to-one-lessness of f[itex]^{-1}[/itex]. Inspite of f[itex]^{-1}[/itex] not being one-to-one, it is turning out to be true while 2 requires any f to be one-to-one.
Now you may ask me as to why f[itex]^{-1}[/itex] is not one-to-one. Let B1 and B2, subsets of Y, be such that range of X under f is a subset of both B1 and B2. Also, let B1[itex]\neq[/itex]B2. But both of them map to X under f[itex]^{-1}[/itex]. Please note that range of f[itex]^{-1}[/itex] is defined to be a set while f, according to Halmos's convention, can operate on a set OR on elements and hence, can have a range whose elements are sets or elements.
In this context, range of f is a set of sets as it is operating on sets. Strictly speaking, we are looking here at f operating on P(X) to P(Y) while f[itex]^{-1}[/itex] operates on P(Y) to P(X).
Kindly let me know as to why 1 is true inspite of the fact that f[itex]^{-1}[/itex] is not one-to-one.