# Is the lower limit topology finer than the standard topology on $\mathbb{R}$?

Is the lower limit topology finer than the standard topology on $\mathbb{R}$?

In Lemma 13.4 on p.82 of Munkres' Topology (2nd ed.), it is stated that the lower limit topology is (strictly) finer than the standard topology on $\mathbb{R}$. In the argument, he is using that the interval $[a,b)$ lies in the interval $(a,b)$ which is certainly not true. On the other hand, the converse is true, that is: $$(a,b) \text{ lies in the interval } [a,b).$$

So we can conclude that the standard topology is finer than the lower limit topology. Am I right? If not then why? I think it's an errata. I have checked some existing errata online but it's not included, though.

• More sequences converge, since having a limit implies having a lower limit. – Adam Hughes Jul 8 '14 at 0:23
• From Topology, a first course (1st edition) by James R. Munkres, on page 82: "Given a basis element $(a,b)$ for $\cal T$, and a point $x$ of $(a,b)$, the basis element $[x,b)$ for $\cal T^\prime$ contains $x$ and lies in $(a,b)$." So he does not claim that $[a,b)$ is contained in $(a,b)$. – user940 Jul 8 '14 at 0:32

Yes! Since one have that $$(a,b) = \cup_{n\ge 1} \ [a+\frac{\epsilon}{n},b)$$ where $\epsilon < \frac{b-a}{2}$.

Note that if for topology ${\mathcal T}_1$ with basis ${\mathcal S}_1$ and topology ${\mathcal T}_2$, one have that ${\mathcal S}_1 \subseteq {\mathcal T}_2$, then ${\mathcal T}_2$ is finer than ${\mathcal T}_1$. In this case if ${\mathcal S}_2$ be a basis for topology ${\mathcal T}_2$ and ${\mathcal S}_2 \not\subseteq{\mathcal T}_1$, then ${\mathcal T}_2$ is strictly finer than ${\mathcal T}_1$.For this, it is enough to note that $[0,1)$ is not open in standard topology.

For more details you can consult this textbook: James Munkres, "Topology; A First Course".

• This isn't quite right. You could correct it by using the intersection rather than union or by changing $-$ to $+$. No matter how big or small $\epsilon$ is then $[a-\frac{\epsilon}{n},b) \supset (a,b)$ and this is a proper inclusion. – Squirtle Aug 12 '15 at 16:24

Well, I don't have a copy of Munkres' book at hand but I doubt that it is said.

If $[a,b)$ is open then $(a,b)=\bigcup_{n\in\mathbb{N}}\left[ a+\frac{1}{n},b\right)$ must be open.

Conversely, $[0,1)$ is not open in the standard topology.

This means that the topology in $\mathbb{R}_l$ is finer that the topology on $\mathbb{R}.$