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Doesn't that mean the interval from 3 to 3, not including 3?provide further context

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The statement that \(\displaystyle (a,b)\) is an open interval in the set of real numbers means that \(\displaystyle a~\&~b\) are two real numbers and \(\displaystyle a<b\).Doesn't that mean the interval from 3 to 3, not including 3?

\(\displaystyle (a,b)=\{x\in\mathbb{R}: a<x<b\}\)

and thus \(\displaystyle (a,a) = \{x \in \mathbb{R} : a < x < a\} = \emptyset\)The statement that \(\displaystyle (a,b)\) is an open interval in the set of real numbers means that \(\displaystyle a~\&~b\) are two real numbers and \(\displaystyle a<b\).

\(\displaystyle (a,b)=\{x\in\mathbb{R}: a<x<b\}\)

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Anyone who can count knows that \(\displaystyle a=a\) so \(\displaystyle (a,a)\) does have two real numbers therefore by definition is not notation for an open interval.and thus \(\displaystyle (a,a) = \{x \in \mathbb{R} : a < x < a\} = \emptyset\)

But [a,a] is.Anyone who can count knows that \(\displaystyle a=a\) so \(\displaystyle (a,a)\) does have two real numbers therefore by definition is not notation for an open interval.

no, that's a closed interval. It contains it's single limit point aBut [a,a] is.