complement, open, close question

Oct 2010
47
0
Support f:\(\displaystyle R^n-> R^m\) and A (subset) \(\displaystyle R^m\). Define a subset of \(\displaystyle R^n\) the pre-image of the set A by\(\displaystyle f"\) and denoted \(\displaystyle \left(f^{-1}(A)\right)\) by

\(\displaystyle \left(f^{-1}(A)\right)\)= {\(\displaystyle x in R^n| f(x) in A\)}

Prove that \(\displaystyle \left(f^{-1}(A)\right)^c\) =\(\displaystyle \left(f^{-1}(A^c)\right)\) , where \(\displaystyle X^c\) denotes the complement of X in\(\displaystyle R^n\).

Prove the \(\displaystyle f:R^n ->R^m\) is continuous <=> for all open V (subset) \(\displaystyle R^m\), \(\displaystyle \left(f^{-1}(V)\right)\)is open in \(\displaystyle R^n\)

Pls help(Crying)
 
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Oct 2010
47
0
I know that \(\displaystyle f^{-1}( A) is mean f(x) =A\)

for the first one that when I want to prove that \(\displaystyle \left(f^{-1}(A)\right)^c\) =\(\displaystyle \left(f^{-1}(A^c)\right)\)
should I use this idea?

and the \(\displaystyle A^c\) does it mean the thing that not in A?
 
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Plato

MHF Helper
Aug 2006
22,507
8,664
If you expect to have your posts answered, you must learn to post in symbols? You can use LaTeX tags.
[noparse]\(\displaystyle \left(f^{-1}(A)\right)^c\)[/noparse] gives \(\displaystyle \left(f^{-1}(A)\right)^c \)

We cannot help on things we cannot read.
 
Oct 2010
47
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thanks for your remind. I am still thinking how can I post them in symbols.
 
Oct 2010
47
0
I know how to prove that since \(\displaystyle f:R^n ->R^m\) is continuous => for all open V (subset) \(\displaystyle R^m\), \(\displaystyle \left(f^{-1}(V)\right)\)is open in \(\displaystyle R^n\)
but I have no idea how can I show the other way around.
 

Plato

MHF Helper
Aug 2006
22,507
8,664
I know how to prove that since \(\displaystyle f:R^n ->R^m\) is continuous => for all open V (subset) \(\displaystyle R^m\), \(\displaystyle \left(f^{-1}(V)\right)\)is open in \(\displaystyle R^n\)
but I have no idea how can I show the other way around.
What does "other way around" mean?
 
Nov 2010
11
1
did you tried by settings open balls?? note that \(\displaystyle B(f(x_{0}),\epsilon)\) is an open subset of \(\displaystyle R^{m}\) (write up the continuity's definition in orden to understand mostly of what I am saying.)
 
Oct 2010
47
0
the other way around mean \(\displaystyle f:R^n ->R^m\) is continuous <= for all open V (subset) \(\displaystyle R^m\), \(\displaystyle \left(f^{-1}(V)\right)\)is open in \(\displaystyle R^n\)
 
Oct 2010
47
0
did you tried by settings open balls?? note that \(\displaystyle B(f(x_{0}),\epsilon)\) is an open subset of \(\displaystyle R^{m}\) (write up the continuity's definition in orden to understand mostly of what I am saying.)
you are saying since the function is a open set, so there exist \(\displaystyle B(f(x_{0}),\epsilon)\) in \(\displaystyle R^{m}\) then use the open ball to prove that this is continuous?
 

Plato

MHF Helper
Aug 2006
22,507
8,664
the other way around mean \(\displaystyle f:R^n ->R^m\) is continuous <= for all open V (subset) \(\displaystyle R^m\), \(\displaystyle \left(f^{-1}(V)\right)\)is open in \(\displaystyle R^n\)
That reply makes no sense whatsoever.
By definition if \(\displaystyle f\) is continuous and \(\displaystyle O\) is open then \(\displaystyle f^{-1}(O)\) is open.

I thought you meant by the ‘other way around’ the following:
\(\displaystyle f\) is continuous and \(\displaystyle M\) is closed then \(\displaystyle f^{-1}(M)\) is close.

That is a theorem. So what do you mean?