Let's denote "A=(a,b)" and "M=[a,b]" are any arbitrary open and closed interval of "\\mathbb{R}" respectively.
Claim: "A" is open set in "\\mathbb{R}"
Proof:
We will show that every point of "A" is interior point of "A",hence "A" will be open in "\\mathbb{R}" . Consider any arbitrary point "p \\in A" and "r_p>0" , such that "B(p,r_p):=\\{x \\in \\mathbb{R} |\\: |x-p|<r_p\\} \\: (\\clubs)" is the open "r_p" -ball of "p" ,Clearly if we choose suitable small enough "r_p" such "r_p" -ball always exist then "B(p,r_p) \\subset A" ,Hence we are done.
Alternatively, observe that
"A=\\cup_{p \\in A} \\: B(p,r_p) \\: (\\spades)"Since, by the hypothesis,"B(p,r_p)" is open set , therefore union of open set is open.Hence "A" is open.
Now, consider the set "M", we claim that "M" is closed in "\\mathbb{R}" .
Consider the compliment of "M" is "M'=\\mathbb{R} \\backslash M" , if we show "M'" is open in "\\mathbb{R}" we will be done.
Let, "m \\in M' \\implies m \\notin M" ,thus as defined in "(\\clubs)" , "B(m,r_m)" and considering "(\\spades)" implies
Hence, "M'" is open in "\\mathbb{R}" as arbitrary union of open set is open set.Therefore, "M" is closed in "\\mathbb{R}" .
We are done.
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