Since every subset is open, every subset is also closed

metric-spaces

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The answer to the question is given below:

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Could somebody please explain me how do they deduce that "Since every subset is open, every subset is also closed". Is this some kind of definition, property?

The way I see that the sets are both open and closed are:

Let $B=X \backslash A$ then $\forall b \in B \exists \epsilon$ s.t $B(b,\epsilon) \subset B$ . Here we can take $\epsilon \in (b,1]$. Hence $B$ is open and by definition, $A$ is closed.

Best Answer

By definition, a subset is closed if its complement is open. If $U$ is a subset, then $X\setminus U$ is also a subset, thus is open if all subsets are open. As $X\setminus U$ is open, then $U$ is closed. This is true for all subsets $U$.

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