Lemma: Arbitrary union of open set is open set.
Proof:
Let fix and denote an open set for each . Now we will show that
is open. Consider any arbitrary at least one such that which implies that there exist an open ball of radius such that (Since, by our hypothesis every is open ,hence, is open for some .)which implies that
Thus, is interior point of and as is arbitrary, therefore all points of are interior points .Hence is open.
Let us denote open and closed intervals respectively.
Claim: is open set in
Proof:
Let, , is a open ball centered at of radius , Thus, clearly
Therefore, from the above Lemma, implies is open.
Now, consider is the complement of ,thus if we prove that is open then is a closed set in .
Clearly, ,therefore again from the above Lemma is open (since, are both open intervals).Hence, we are done.
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