# complement, open, close question

#### mathbeginner

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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#### mathbeginner

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
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.

#### mathbeginner

thanks for your remind. I am still thinking how can I post them in symbols.

#### mathbeginner

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
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?

#### teachermath

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.)

#### mathbeginner

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 $$\displaystyle R^n$$

#### mathbeginner

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
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?