Answer to Question #115964 in Real Analysis for Sheela John

Question #115964
Prove that an open interval in R is an open set and a closed intervals is a closed set
1
Expert's answer
2020-05-20T20:01:21-0400

For definiteness, let us consider any open interval in "R" is "(c,d)=\\{x\u2208R\u2223c<x<d\\}"

Furthermore, let "a \\in (c,d)" and "\u03f5\u2264a\u2212c\\ and\\ \u03f5\u2264d\u2212a" .

Now "a\u2208(c,d)" and recall that the ϵ-neighborhood of "a" in the set "\\implies x\\in V_\\epsilon (a)" so "a-\\epsilon <x<a+\\epsilon" .

If we take "\u03f5=min\\{a\u2212c,d\u2212a\\}" , then "\\epsilon \u2264a\u2212c\\ and\\ \\epsilon\u2264d\u2212a" .

So, "c=a\u2212(a\u2212c)<a\u2212\u03f5<x<a+\u03f5<a+(d\u2212a)=d"

"\\implies x\u2208(c,d) \u27f9 V_\u03f5(a)\u2286(c,d)" .

Hence, every open interval in "R" is an open set.


Now, let "[c,d]" is closed interval in "R" , so "[c,d]^c = (-\\infin,c) \\cup (d,\\infin)" . In part (i), we proved every open interval is open set and also we known union of two open set is open. So compliment of given closed interval is open set. So, given closed interval is closed set.


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