Find the boundary of a set $A$ defined by $A=(-1,1) \cup \lbrace 2 \rbrace$ in the lower limit topology and in the standart topology.
For find the $\partial(A)$ in standart topology, i remember the definition of it, such as $\partial (A)=Cl(A)-Int(A)$, since $Int(A)$ is the most larguest open in $A$ and $A$ is a open set add a closed set, then $Int(A)=(-1,1)$,so now to find the clasusure we need fisrt find $A^{\prime}$ (Set of limit points), in these case, $2$ can´t be a limit point since exist a neighborhood that contains 2 and not intersect $A$, im claim that $-1,1$ are limit poits since every neighborhood that contains $-1,1$ intersect $A$, let $U$ neighborhood of $-1$ then $-1\in(a,b):a<-1<b$ and since $b+\varepsilon<1$ or $b+\varepsilon>1$ for $\varepsilon$ arbitrary then $U$ intersects $A$, and with the $1$ use a similar argument, so now we use the fact that $Cl(A)=A\cup A^{\prime}$ then we get
$\partial(A)=\lbrace-1,1,2 \rbrace$.
In the lower Limit topology im get that $\partial(A)$ is the same that in the standart topology, im corect, or i am not understanding fine the concepts.
If case of be possibly, someone can help me with this in the lower limit toplogy, and too explain me about of if $[a,b)$ is a basis element in the lower limit topology and $x$ belongs to it. then
is correct claim that $a\leq x <b$. Any help is useful.
Find the boundary of a set $A$ in Lower limit topology
general-topology
Best Answer
Yes, $\{-1,1,2\}$ is the boundary of $A$ in the usual topology. In the lower limit topology, however, $1$ is not in the closure of $A$, so it cannot be in the boundary of $A$: $[1,2)$ is an open nbhd of $1$ in the lower limit topology that does not contain any point of $A$. $-1$, on the other hand, is in the closure of $A$ in the lower limit topology: for each $\epsilon>0$, $[-1,-1+\epsilon)\cap A\ne\varnothing$. $2\in A$, so of course $2$ is in the closure of $A$ in the lower limit topology, and $2$ is not in the interior of $A$ in the lower limit topology, since $A$ does not contain any set of the form $[2,2+\epsilon)$ with $\epsilon>0$, so $2$ is in the boundary of $A$ in the lower limit topology. Finally, it’s easy to check that the points of $\Bbb R\setminus\{-1,1,2\}$ are not in the boundary of $A$ in the lower limit topology, so the boundary of $A$ in the lower limit topology is just $\{-1,2\}$.