The proof in question (from the book "Topology" by Munkres):

Let $\mathcal T$ and $\mathcal T_\mathscr{l}$ be the standard and lower limit topology on $\mathbb R$ respectively. Given a basis element $(a,b)$ for $\mathcal T$ and a point $x\in (a,b)$, the basis element $[x,b)$ for $\mathcal T_\mathscr{l}$ contains $x$ and lies in $(a,b)$. Conversely, given the basis element $[x,d)$ for $\mathcal T_\mathscr{l}$, there is no open interval in $\mathcal T$ that contains $x$ and lies in $[x,d)$. Thus $\mathcal T_\mathscr l$ is strictly finer than $\mathcal T$.

There was a question on it before, but his actual proof was never addressed. I understand the proof involving the union of an infinite collection of sets. However, I don't get how the above proof proves that all elements of the latter topology are in the former, which as I understand is the definition of a finer topology.

  • 1
    $\begingroup$ Consider the analogy of sand in his book. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 29 '18 at 17:04
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    $\begingroup$ @GNUSupporter Honestly, it is passages like the sand analogy that make me really like Munkres as an introductory text. Some of the later chapters lose the thread a little (in my opinion), but the first half of that book is golden. $\endgroup$ – Xander Henderson Mar 29 '18 at 17:06
  • $\begingroup$ The lower limit topology on $\Bbb R$ is also called the Sorgenfrey line, sometimes denoted $\Bbb R_l.$ Its square $\Bbb R^2_l $ is called the Sorgenfrey plane. $\endgroup$ – DanielWainfleet Mar 30 '18 at 4:57

Of $O\in\mathcal{T}$, then $O$ can be written as the union of intervals of the type $(a,b)$. Each interval $(a,b)$, in turn, can be written as$$\bigcup_{x\in(a,b)}[x,b),$$which belongs to $\mathcal{T}_1$. Therefore, $O\in\mathcal{T}_1$.

  • $\begingroup$ But isn't Munkres missing this step? I don't see how he explicitly shows that the interval (a,b) can also be found in $\mathcal T_\mathscr l$ $\endgroup$ – oddic Mar 29 '18 at 17:13
  • $\begingroup$ @oddic He proves that, for each $x\in(a,b)$, there is an element $S\in\mathcal{T}_1$ such that $x\in S$ and that $S\subset(a,b)$. I suppose that he didn't feel the need to deduce from this that $(a,b)\in\mathcal{T}_1$. $\endgroup$ – José Carlos Santos Mar 29 '18 at 17:16
  • $\begingroup$ But doesn't the proof require showing that there is an element $S\in \mathcal T_\mathscr l$ which is a union of infinite basis elements that is equal to $(a,b)$, and not just a subset? I'm sorry if I'm not understanding correctly $\endgroup$ – oddic Mar 29 '18 at 17:34
  • $\begingroup$ @oddic If, for each $x\in(a,b)$, there is a $S_x\in\mathcal{T}_1$ such that $x\in S_x$ and that $S_x\subset(a,b)$, then$$(a,b)=\bigcup_{x\in(a,b)}S_x\in\mathcal{T}_1.$$ $\endgroup$ – José Carlos Santos Mar 29 '18 at 17:43
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    $\begingroup$ @oddic In Lemma 13.3 (p.81) the author already gave an equivalent condition of a topology being finer than the other topology. So in the following Lemma 13.4 he did not need to mention it again, I think. $\endgroup$ – ChoF Mar 30 '18 at 5:12

What Munkres is doing:

If $T_1,T_2$ are topologies on a set $R,$ and $B_1,B_2$ are bases for $T_1,T_2$ respectively, then to show that $T_1\subset T_2,$ it suffices to show that whenever $x\in b_1\in B_1$ there exists $b_2\in B_2$ with $x\in b_2\subset b_1.$

That implies that every $b_1\in B_1$ is a union of members of $B_2,$ so $b_1\in T_2$. So $B_1\subset T_2,$ so any $t\in T_1,$ being a union of members of $B_1,$ is a union of members of $T_2,$ so $t\in T_2.$

In this case we have $R=\Bbb R$ and $T_1=\mathcal T$ and $T_2=\mathcal {T_l}$ and $B_1=\{(a,b): a,b\in \Bbb R\}$ and $B_2=\{[x,b):x,b\in \Bbb R\}.$

Therefore $\mathcal T\subset \mathcal T_l.$

(And, obviously, they are not equal, because $[0,1)\in \mathcal T_l$ \ $\mathcal T.$)

  • $\begingroup$ Thanks, I understand what I was missing now! $\endgroup$ – oddic Apr 1 '18 at 0:02
  • $\begingroup$ In many cases the use of particular bases simplifies things. Not surprising when you consider that usually what you do is show that if something holds for a certain subset of a topology (the base/basis) then it holds for the whole topology..... Many write "base", not "basis". I prefer "base", especially in the context of topological vector spaces, where "basis" also has vectorial meanings. $\endgroup$ – DanielWainfleet Apr 1 '18 at 16:37

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