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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?

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2 Answers 2

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.

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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.

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