# Continous function in topologies

• Apr 27th 2010, 11:32 AM
Amer
Continous function in topologies
$f: (R,T_u) \rightarrow (R,T_{cof})$

defined by

$f(x)=\left\{ \begin {array}{cc} -1 ,& \mbox x\in Q \\ 1 , &\mbox x\in Q^c \end {array}\right.$

is this function continuous, and is this function closed

I found it closed and not continuous

am I right ?

• Apr 27th 2010, 11:52 AM
Drexel28
Quote:

Originally Posted by Amer
$f: (R,T_u) \rightarrow (R,T_{cof})$

defined by

$f(x)=\left\{ \begin {array}{cc} -1 ,& \mbox x\in Q \\ 1 , &\mbox x\in Q^c \end {array}\right.$

is this function continuous, and is this function closed

I found it closed and not continuous

am I right ?

What are $T_u$ and $T_{\text{cof}}$? The usual and cofinite topologies? And I assume $R=\mathbb{R}$.

I agree that it's closed. If $U\subseteq\mathbb{R}$ then $f(U)\subseteq\{-1,1\}$ and so $\mathbb{R}-f(U)\supseteq \mathbb{R}-\{1,-1\}$ is open. The image of any subset of $\mathbb{R}$ is closed.

I also agree that it can't be continuous. For, if $f:\left(\mathbb{R},T_u\right)\to\left(\mathbb{R},T _{\text{cof}}\right)$ were continuous then $f:\left(\mathbb{R},T_u\right)\to\left(\{-1,1\},T_s\right)$ where $T_s$ is the subspace topology would be continuous. But, every subspace of a cofinite space has the cofinite topology, and a finite cofinite space is discrete. And so, $\left(\{-1,1\},T_s\right)\approx D$ where $D$ is the two point discrete space.

Thus, if $f$ were continuous there would be a map $\varphi:\left(\mathbb{R},T_u\right)\to D$ which is surjective and continuous. What's the problem with that?
• Apr 27th 2010, 12:05 PM
Amer
Yeah it is the usual and cofinite topologies and the set is the real numbers
but I do not know how to write R with two lines

$\varphi : (\mathbb{R} ,T_u) \rightarrow D$

suppose D= {1,2}

$\varphi ^{-1} (1)$ and $\varphi ^{-1} (2)$ open sets in $\mathbb{R}$

and $\varphi ^{-1} (1) \cup \varphi ^{-1} (2) = \mathbb{R}$

that means real numbers with usual topology is connected contradiction
• Apr 27th 2010, 12:08 PM
Drexel28
Quote:

Originally Posted by Amer
that means real numbers with usual topology is $\bf{not}$ connected contradiction

Exactement.