Take the tour ×  In Munkres' "Topology", Section 18, Example 3
(pg. 104), it is stated that the identity function  $$
g:\mathbb{R}_l\rightarrow\mathbb{R}, g(x)=x $$  where
$\mathbb{R}$ has the usual topology and $\mathbb{R}_l$ has the
lower limit topology, is continuous, because the inverse image
of an open $(a,b)$ of $\mathbb{R}$ is the open set $(a,b)$ in
$\mathbb{R}_l$. But the basis of the $\mathbb{R}_l$ topology is
formed by sets on the form $[c,d)$.  I figured out that $(a,b)$
must be open in $\mathbb{R}_l$ since $$
(a,b)=\bigcup_{i=1}^{\infty}\left.[a-\frac{1}{n},b \right.) $$
Can someone tell me if this is correct?  share|improve this
question add comment  2 Answers  up vote 2 down vote accepted
That would be fine if you wrote:  $$(a,b) =
\bigcup_{i=1}^{\infty}[a+\frac{1}{n},b).$$  This way $a$ is not
in that union.  Anyway, all you need to check is that the pre-
image of any basis element is open. The above demonstrates that
indeed, the pre-image is open because you expressed it as a
union of open se