0
$\begingroup$

Can anyone tell me what open sets will look like for say $(0,1),[0,1),(0,1],[0,1]$ in the lower limit topology? I get what they will look like in the usual topology, but I am not sure what they will look like in the $\mathbb{R}_l$.

My guess would be, for $(0,1)_{l}$ open sets would look like $[a,b)$ where $a\ne 0,$ and $b\le 1$ that is $0<a<b\le1?$ Would appreciate the help.

$\endgroup$
  • 1
    $\begingroup$ For $(0,1)_l$, the open set (in fact basic elements) are of the form $(0,b)$ for $0<b<1$ and $[a,b)$ for $0<a<b<1$ and $[a,1)$ also $\endgroup$ – Qurultay Mar 26 '18 at 13:14
  • $\begingroup$ $(0,1)=\bigcup_n[1/n,1)$, $[0,1)$ is open by definition. $[0,1]$ is closed, since it is the complement of the open set $\bigcup_n[-n,0)\cup[1-1/n,n)$. $\endgroup$ – user545497 Mar 26 '18 at 13:16
  • $\begingroup$ $(0,1]$ and $[0,1]$ cannot be open because every neighborhood of $1$ contains points outside of them. $(0,1]$ cannot be closed because $1/n\to 0\notin (0,1]$. $\endgroup$ – user545497 Mar 26 '18 at 13:22
  • $\begingroup$ Thank you for the help. Could anyone also help me with the remaining $3?$ For $[0,1)_l,$ will the open sets look like $(0,b),[a,b)$ for $0\le a <b\le 1?$ $\endgroup$ – Aurora Borealis Mar 26 '18 at 13:29
  • $\begingroup$ (0,b) is also open within lower limit (0,1). $\endgroup$ – William Elliot Mar 26 '18 at 21:20
1
$\begingroup$

Partitions and Equivalence Relations. A partition of a set $S$ is a family $P$ of pair-wise disjoint subsets of $S$ whose union is $S.$ An equivalence relation on $S$ is a binary relation $\sim$ on $S$ which is

(i) Reflexive: $\forall x\in S\;(x\sim x)$

(ii) Symmetric: $\forall x,y \in S\;(x\sim y\iff y\sim x) $

(iii) Transitive: $\forall x,y,z \in S\;((x\sim y \land y\sim z)\implies x\sim z.)$

Every partition on $S$ determines an equivalence relation on $S,$ and vice-versa: If $P$ is a partition of $S$ then for $x,y\in S$ let $x\sim y$ iff $x,y$ belong to the same member of $P.$ If $\sim$ is an equivalence relation on $S,$ then for each $x\in S$ let $[x]_{\sim}=\{y\in S: y\sim x\}.$ Then $P=\{[x]_{\sim}: x\in S\}$ is a partition of $S.$ (Note: $[x]_{\sim}$ is called an equivalence class.)

Notation: For $x,y \in \Bbb R$ let In$[x,y]=[x,y]\cup [y,x].$ That is, in the standard topology on $\Bbb R$ the set In$[x,y]$ is the closed interval from $x$ to $y$ (and vice-versa).

Let $S$ be open in the lower-limit topology. For $x,y\in S$ let $x\sim y \iff$ In$[x,y]\subset S.$ It is easily shown that

(i'). $\sim$ is an equivalence-relation on $S.$

(ii'). For $x\in S$ the equivalence-class $[x]_{\sim}=\{y\in S :y\sim x\}$ is a convex set of non-zero length.

(iii'). If $[x]_{\sim}$ is bounded above then it does not contain its sup, but if it is bounded below then it may or may not contain its inf .

The set $P$ of equivalence classes is a partition of $S$, so the equivalence-classes are pair-wise disjoint. Each equivalence class contains a rational, and the classes are pair-wise disjoint, so $P$ is finite or countably infinite.

And of course $S=\cup P.$

Observe that if $z=\sup \;[x]_{\sim}<\infty$ then $z\not \in S .$

Observe that if $z=\inf \;[x]_{\sim}\in S$ then $z\in [x]_{\sim}$ and $z=\sup\; (\; (-\infty, z)\setminus S\;).$

Examples. (1). Let $S=[0,1)$ or $S=(0,1).$ Then $P=\{S\}.$

(2).Let $C$ be the Cantor set. Let $[0,1]\setminus C =\cup \{(a_n,b_n):n\in \Bbb N\}$ where $(a_n,b_n)\cap (a_m,b_m)=\emptyset$ when $n\ne m.$ Let $S=\{[a_n,b_n): n\in \Bbb N\}.$ Each $[a_n,b_n)$ belongs to $P.$ Just as in the standard topology, open sets in the lower-limit topology can be "complicated".

$\endgroup$
  • $\begingroup$ Ok this is alittle more than what I can absorb, I will try to understand this, thank you for the help. $\endgroup$ – Aurora Borealis Mar 27 '18 at 7:36
  • $\begingroup$ OK. Partitions and equivalence relations are a useful & widely used tool. $\endgroup$ – DanielWainfleet Mar 27 '18 at 18:48

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.