Let's denote and are any arbitrary open and closed interval of respectively.
Claim: is open set in
Proof:
We will show that every point of is interior point of ,hence will be open in . Consider any arbitrary point and , such that is the open -ball of ,Clearly if we choose suitable small enough such -ball always exist then ,Hence we are done.
Alternatively, observe that
Since, by the hypothesis, is open set , therefore union of open set is open.Hence is open.
Now, consider the set , we claim that is closed in .
Consider the compliment of is , if we show is open in we will be done.
Let, ,thus as defined in , and considering implies
Hence, is open in as arbitrary union of open set is open set.Therefore, is closed in .
We are done.
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