1
$\begingroup$

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.

$\endgroup$
  • 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
  • 2
    $\begingroup$ @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
1
$\begingroup$

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, $O\in\mathcal{T}_1$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
0
$\begingroup$

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 $[0,1)\in \mathcal T_l$ \ $\mathcal T.$)

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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