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In Sorgenfrey line, evaluate the following and prove your result



and prove result

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Yes I have tried – user84515 Jul 21 '13 at 21:41

HINTS: Let $A$ be a subset of $\Bbb S$, the Sorgenfrey line. A point $x$ is in the interior of $A$ if and only if there is a $y>x$ such that $[x,y)\subseteq A$. You’re interested in the set $A=(-2,-1]\cup\{0\}\cup[1,2)$. I’ll get you started. Consider the point $0$: no matter what $y>0$ you choose, $[0,y)$ is not a subset of $A$, so $0\notin\operatorname{int}A$. For most of the other points $x\in A$ you should have little trouble finding a $y>x$ such that $[x,y)\subseteq A$; in fact, there’s only one other point of $A$ that is not in the interior of $A$.

Similarly, $x\in\operatorname{cl}A$ if and only if every nbhd $[x,y)$ of $x$ contains a point of $A$. Certainly this is true if $x\in A$. What points $x\in\Bbb S\setminus A$ have nbhds of the form $[x,y)$ such that $[x,y)\cap A=\varnothing$? Those points are not in the closure of $A$. (The points that you should focus on in particular are $-2$ and $2$; can you see why?)

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I know the definitions of interior and closure but idon't know how to calculate and proof – user84515 Jul 21 '13 at 21:19
@user84515: Then you should have absolutely no trouble answering the question, especially with the additional pointers that I gave. – Brian M. Scott Jul 21 '13 at 21:21

Lets begin with $\operatorname{cl}(A\cup B\cup c)=\operatorname{cl}(A)\cup \operatorname{cl}(B)\cup \operatorname{cl}(C)$.

while $ A= (-2,-1]$,$ B=\{0\}$ and $ C= [1,2)$.

Because Sorgenfrey Topology contains $\mathbb R$ with the Standard Topology -> [a,b] is closed on Sorgenfrey.

$\operatorname{cl}(C)$ ={S|S is the minimal closed set that contains A}=[1,2]


lets find $X-A$ :$X-A=(-\infty,-2] \cup\ (-1,\infty)$ , $X-A$ is not open so A is not closed than :


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interior? and proof – user84515 Jul 21 '13 at 21:37
the interior is the maximal set in the original set and remember that Sorgenfrey contains the standard topology so it also contains its open sets – Eli Elizirov Jul 21 '13 at 21:40

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