# Topological Spaces

• Mar 4th 2010, 03:49 AM
slevvio
Topological Spaces
Let $\displaystyle \mathcal{Z}$ be the family of sets $\displaystyle U$ in $\displaystyle \mathbb{R}$ such that $\displaystyle U$ is empy or $\displaystyle \mathbb{R} \setminus U$ is finite.
Show that $\displaystyle \mathcal{Z}$ is a topology on $\displaystyle \mathbb{R}$. By considering the inverse images of closed sets, show that every polynomial function $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ is $\displaystyle (\mathcal{Z},\mathcal{Z})$-continuous.

Hey everyone, I was wondering if anyone could assist me with the 2nd part. If $\displaystyle A$ is closed in $\displaystyle \mathbb{R}$ then $\displaystyle \mathbb{R} \setminus A$ is the empty set, or A is finite, i.e. A is $\displaystyle \mathbb{R}$ or finite.

I don't understand though how $\displaystyle f^{-1} (A)$ is closed and I would really appreciate some advice here, thank you so much. This would show continuity of the polynomial functions.
• Mar 4th 2010, 04:40 AM
tonio
Quote:

Originally Posted by slevvio
Let $\displaystyle \mathcal{Z}$ be the family of sets $\displaystyle U$ in $\displaystyle \mathbb{R}$ such that $\displaystyle U$ is empy or $\displaystyle R \setminus U$ is finite.
Show that $\displaystyle \mathcal{Z}$ is a topology on $\displaystyle \mathbb{R}$. By considering the inverse images of closed sets, show that every polynomial function $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ is $\displaystyle (\mathcal{Z},\mathcal{Z})$-continuous.

Hey everyone, I was wondering if anyone could assist me with the 2nd part. If $\displaystyle A$ is closed in $\displaystyle \mathbb{R}$ then $\displaystyle \mathbb{R} \setminus A$ is the empty set, or A is finite, i.e. A is $\displaystyle \mathbb{R}$ or finite.

I don't understand though how $\displaystyle f^{-1} (A)$ is closed and I would really appreciate some advice here, thank you so much. This would show continuity of the polynomial functions.

Hint: if f(x) is a polynomial of degree > 0, how many solutions at most can the equation $\displaystyle f(x)=k$ , for fixed k, have?

So now, how many elements can the set $\displaystyle f^{-1}(A)$ have, when A is finite?

Tonio
• Mar 4th 2010, 05:37 AM
slevvio
it can have at most the degree of the polynomial am I right? eg a polynomial of degree three has 3 complex roots so at most 3 real roots.

so then the number of elements in $\displaystyle f^{-1} (A) \le deg(f)|A|$. Is this correct?

i.e. $\displaystyle |f^{-1} (A)| \le deg(f)|A| < \infty$ Hence $\displaystyle |f^{-1} (A)|$ is finite!

But what about if A = $\displaystyle \mathbb{R}$? Then I guess I am trying to show that $\displaystyle f^{-1}(\mathbb{R}) = \{ x \in \mathbb{R} : f(x) \in \mathbb{R} \} = \mathbb{R}$. I guess if f is a polynomial it just moves every point in R to some other point in R. i.e. take any point in R and the polynomial will plop out another point, so it is its own image.

Thanks for the help
• Mar 4th 2010, 09:01 AM
Drexel28
Quote:

Originally Posted by slevvio
Let $\displaystyle \mathcal{Z}$ be the family of sets $\displaystyle U$ in $\displaystyle \mathbb{R}$ such that $\displaystyle U$ is empy or $\displaystyle \mathbb{R} \setminus U$ is finite.
Show that $\displaystyle \mathcal{Z}$ is a topology on $\displaystyle \mathbb{R}$. By considering the inverse images of closed sets, show that every polynomial function $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ is $\displaystyle (\mathcal{Z},\mathcal{Z})$-continuous.

Hey everyone, I was wondering if anyone could assist me with the 2nd part. If $\displaystyle A$ is closed in $\displaystyle \mathbb{R}$ then $\displaystyle \mathbb{R} \setminus A$ is the empty set, or A is finite, i.e. A is $\displaystyle \mathbb{R}$ or finite.

I don't understand though how $\displaystyle f^{-1} (A)$ is closed and I would really appreciate some advice here, thank you so much. This would show continuity of the polynomial functions.

This is called the cofinite topology on $\displaystyle \mathbb{R}$. Notice that the resulting topological space is $\displaystyle T_1$. And so let $\displaystyle p:\mathbb{R}\mapsto\mathbb{R}$ be a polynomial, and let $\displaystyle F\subseteq\mathbb{R}$ be closed, then since it is the compliment of an open set we have that $\displaystyle F$ is finite. Let $\displaystyle \deg\text{ }p=n$. Then clearly for each $\displaystyle \xi\in F$ we have that $\displaystyle \text{card }p_n^{-1}(\{\xi\})\leqslant n$ (otherwise the polynomial $\displaystyle p_n(x)-\xi$ would be a polynomial of degree $\displaystyle n$ with more than $\displaystyle n$ roots! Thus, we see that $\displaystyle \text{card }p^{-1}(F)=\text{card }\bigcup_{\xi\in F}p^{-1}(\xi)\leqslant n\cdot\text{card }F$. Thus, $\displaystyle \text{card }p^{-1}(F)$ is finite and thus by $\displaystyle T_1$ness closed.

The conclusion follows.