complement topology

Question

Prove whether or not each given function is continuous

a)$f:\mathbb{R}_l \rightarrow \mathbb{R}$ is define by $f(x)=3x-5$ where $\mathbb{R}_l$ is the real line with lower limit topology.

b)$g:\mathbb{R}_{fc} \rightarrow \mathbb{R}$ defined by $g(x)=3x-5$ where $\mathbb{R}_{fc}$ is the real line with finite complement topology.

For part a), suppose $(a,b) \subset \mathbb{R}$. Then $f^{-1}((a,b))=(\frac{a+5}{3},\frac{b+5}{3})$. Note that $\lbrack \frac{a+5}{6},\frac{b+5}{6}) \subset (\frac{a+5}{3},\frac{b+5}{3})$. Hence, $f$ is continuous.

For part b), I know that $g$ is not continuous because for any preimage of open set, there is no open sets in finite complement topology which is a subset of open set. But I don't know how to show this rigorously. Can anyone guide me?

Answer 1

The fact that $$\left[\frac{a+5}6,\frac{b+5}6\right)\subseteq\left(\frac{a+5}3,\frac{b+5}3\right)$$ is irrelevant; in order to show that $f$ is continuous, you must show that $f^{-1}[(a,b)]$ is open in $\Bbb R_\ell$. As you say, this inverse image is $\left(\frac{a+5}3,\frac{b+5}3\right)$, so you need to show that $\left(\frac{a+5}3,\frac{b+5}3\right)$ is open in $\Bbb R_\ell$. To do so, just show that $(x,y)$ is open in $\Bbb R_\ell$ for any $x,y\in\Bbb R$ with $x<y$. (That shows, by the way, that every open set in $\Bbb R$ is also open in $\Bbb R_\ell$.)

For (b), consider the open interval $(1,4)$ in $\Bbb R$: $g^{-1}[(1,4)]=(2,3)$. $\Bbb R\setminus(2,3)$ is clearly not finite, since it includes (for instance) the interval $(0,1)$, so $(2,3)$ is not open in $\Bbb R_{fc}$, and $g$ is therefore not continuous.