Show that $f$ is continuous in the relative topology

Question

Let $(X, T)$ be the subspace of $\mathbb R$ given by $X = [0,1]\cup [2,4]$. Define $f: (X,T) \to \mathbb R$ by $$f(x)= \begin{cases}1 & x \in [0,1] \\ 2 & x \in [2,4]\end{cases}$$ Show that $f$ is continuous.

Let $A= \{1\} \subset \mathbb R$ then clearly $A$ is closed in $\mathbb R$ with the usual topology. Now $f^{-1}(A)= [0,1]$. I need to show that $[0,1]$ is closed in $X$ with the relative topology on $X$.

Similarly let $B= \{2\} \subset \mathbb R$ then $B$ is closed in $\mathbb R$ with the usual topology. Now $f^{-1}(B)= [2,4]$. I need to show that $[2,4]$ is closed in $X$ with the relative topology on $X$.

This then shows that the preimage of closed sets under $f$ is closed, and hence $f$ is continuous.

I am, however, not sure how to show that $[0,1]$ and $[0,2]$ are closed in the relative topology of $X$.

Answer 1

Let $U\subseteq \mathbb{R}$ an open set. We show that $f^{-1}(U)$ in an open set in $X$. If $1,2\not\in U$ we have $f^{-1}(U)=\emptyset$ that is open. If $1\in U$ but $2\not\in U$ we have $f^{-1}(U)=[0,1]$ that is open in the subspace topology of $X$, in fact you can take the open interval $(-\epsilon,1+\epsilon)\supseteq[0,1]$ for some $\epsilon>0$. The case $1\not\in U$ but $2\in U$ is analogue. If $1,2\in U$ we have $f^{-1}(U)=X$ that is open.

Answer 2

Recall: the closed (resp. open) sets in the relative topology are of teh form $C\cap X$, where in this case $C$ is a closed (resp. open)set of the "whole space" (in this case, the real numbers). Now, how do you see the problem? ;-)

Answer 3

You can also use the following fact : Let $X\subset Y$, if $X$ is closed in $Y$ then, every closed set in $X$ is closed in $Y$.