Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question
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 sets, which is again open.

share|cite|improve this answer
Signal error! Thanks! – Marra Jan 13 '13 at 12:55

This is correct. Indeed if you read the wikipedia article this showed $R_{l}$ is finer than $R$ under usual topology.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.