# Thread: complement, open, close question

1. ## complement, open, close question

[QUOTE]Support f: $R^n-> R^m$ and A (subset) $R^m$. Define a subset of $R^n$ the pre-image of the set A by $f"$ and denoted $\left(f^{-1}(A)\right)$ by

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

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

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

Pls help

2. I know that $f^{-1}( A) is mean f(x) =A$

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

and the $A^c$ does it mean the thing that not in A?

3. If you expect to have your posts answered, you must learn to post in symbols? You can use LaTeX tags.
$$\left(f^{-1}(A)\right)^c$$ gives $\left(f^{-1}(A)\right)^c$

We cannot help on things we cannot read.

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

5. I know how to prove that since $f:R^n ->R^m$ is continuous => for all open V (subset) $R^m$, $\left(f^{-1}(V)\right)$is open in $R^n$
but I have no idea how can I show the other way around.

6. Originally Posted by mathbeginner
I know how to prove that since $f:R^n ->R^m$ is continuous => for all open V (subset) $R^m$, $\left(f^{-1}(V)\right)$is open in $R^n$
but I have no idea how can I show the other way around.
What does "other way around" mean?

7. did you tried by settings open balls?? note that $B(f(x_{0}),\epsilon)$ is an open subset of $R^{m}$ (write up the continuity's definition in orden to understand mostly of what I am saying.)

8. the other way around mean $f:R^n ->R^m$ is continuous <= for all open V (subset) $R^m$, $\left(f^{-1}(V)\right)$is open in $R^n$

9. Originally Posted by teachermath
did you tried by settings open balls?? note that $B(f(x_{0}),\epsilon)$ is an open subset of $R^{m}$ (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 $B(f(x_{0}),\epsilon)$ in $R^{m}$ then use the open ball to prove that this is continuous?

10. Originally Posted by mathbeginner
the other way around mean $f:R^n ->R^m$ is continuous <= for all open V (subset) $R^m$, $\left(f^{-1}(V)\right)$is open in $R^n$
That reply makes no sense whatsoever.
By definition if $f$ is continuous and $O$ is open then $f^{-1}(O)$ is open.

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

That is a theorem. So what do you mean?

11. Originally Posted by mathbeginner
you are saying since the function is a open set, so there exist $B(f(x_{0}),\epsilon)$ in $R^{m}$ then use the open ball to prove that this is continuous?
exactly my friend.

this demostration is strong enought to remain writable even in topological settings without assumptions about metric stuff.

12. Originally Posted by Plato
That reply makes no sense whatsoever.
By definition if $f$ is continuous and $O$ is open then $f^{-1}(O)$ is open.

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

That is a theorem. So what do you mean?
the equation is $f:R^n ->R^m$ is continuous <=> for all open V (subset) $R^m$, $\left(f^{-1}(V)\right)$is open in $R^n$.

since it is <=> so after proving => then I have to prove <=
$f:R^n ->R^m$ is continuous <= for all open V (subset) $R^m$, $\left(f^{-1}(V)\right)$is open in $R^n$.
I did prove => but I have no idea how to prove <=

13. Originally Posted by mathbeginner
the equation is $f:R^n ->R^m$ is continuous <=> for all open V (subset) $R^m$, $\left(f^{-1}(V)\right)$is open in $R^n$.

since it is <=> so after proving => then I have to prove <=
$f:R^n ->R^m$ is continuous <= for all open V (subset) $R^m$, $\left(f^{-1}(V)\right)$is open in $R^n$.
I did prove => but I have no idea how to prove <=
You need merely note that if the preimage of open sets is open then for each $B_{\varepsilon}(f(x))$ you have that $f^{-1}\left(B_{\varepsilon}(f(x))\right)$ is open and so there is some $B_{\delta}(x)\subseteq f^{-1}\left(B_{\varepsilon}(f(x))\right)$ and so...

14. Hi again ^^

I think here a more metric proof is needed, because of we are in $\mathbb{R}^{m}$ so I prefer to say that:

Let $x_{0}\in\mathbb{R}^{n}$be fixed and let $\epsilon$ be a positive number, we have that $B(f(x_{0}),\epsilon)$ is an open set of $\mathbb{R}^{m}$, so by hypothesis $f^{-1}(B(f(x_{0}),\epsilon))$ should be an open set in $\mathbb{R}^{n}$.
Now we take $x\inf^{-1}(B(f(x_{0}),\epsilon))$, which means that $f(x)\in B(f(x_{0}),\epsilon)$ and therefore $|f(x)-f(x_{0})|<\epsilon$.
On the other hand you'll se that there is a positive number $\delta$ such that $|x-x_{0}|<\delta$ (look at the set where $x$ lies).
Finally we see continuity of $f$ at $x$.