Topological Spaces

**slevvio** · March 4th 2010, 03:49 AM

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

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

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

**tonio** · March 4th 2010, 04:40 AM

Originally Posted by slevvio

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

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

I don't understand though how $f^{-1} (A)$ 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 $f(x)=k$ , for fixed k, have?

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

Tonio

**slevvio** · March 4th 2010, 05:37 AM

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 $f^{-1} (A) \le deg(f)|A|$ . Is this correct?

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

But what about if A = $\mathbb{R}$ ? Then I guess I am trying to show that $f^{-1}(\mathbb{R}) = \{ x \in \mathbb{R} : f(x) \in \mathbb{R} \} = \mathbb{R}$ . 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

**Drexel28** · March 4th 2010, 09:01 AM

Originally Posted by slevvio

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

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

I don't understand though how $f^{-1} (A)$ 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 $\mathbb{R}$ . Notice that the resulting topological space is $T_1$ . And so let $p:\mathbb{R}\mapsto\mathbb{R}$ be a polynomial, and let $F\subseteq\mathbb{R}$ be closed, then since it is the compliment of an open set we have that $F$ is finite. Let $\deg\text{ }p=n$ . Then clearly for each $\xi\in F$ we have that $\text{card }p_n^{-1}(\{\xi\})\leqslant n$ (otherwise the polynomial $p_n(x)-\xi$ would be a polynomial of degree $n$ with more than $n$ roots! Thus, we see that $\text{card }p^{-1}(F)=\text{card }\bigcup_{\xi\in F}p^{-1}(\xi)\leqslant n\cdot\text{card }F$ . Thus, $\text{card }p^{-1}(F)$ is finite and thus by $T_1$ ness closed.

The conclusion follows.

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