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Prove that the intervals of the form $[a,b)$ are closed in the lower limit topology on $\mathbb{R}$.

I manage to prove the statement by considering its complement, which is $\mathbb{R} \backslash [a,b)$. Then express it as union of open sets in lower limit topology.

My question is why $[a,b)$ is closed intuitively? Because when I first look at this, the interval $[a,b)$ is open in the lower limit topology.

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To me the fact that the complement is pretty obviously open in the lower limit topology is an intuitive reason for the interval’s being closed. – Brian M. Scott Oct 28 '13 at 8:16
@kahen: I just wish that the actor had spoken as clearly as Hitler did in his speeches — I caught only tiny bits of what he was actually saying! – Brian M. Scott Oct 28 '13 at 8:32
@BrianM.Scott: so in this topology, open sets are also closed sets and closed sets are also open sets? – Idonknow Oct 28 '13 at 8:34
Not in general, no. Sets of the form $[a,b)$ are both closed and open, as are finite unions of them, but $(0,1)$, for instance, is a set that is open but not closed, and $\Bbb Z$ is a set that is closed but not open. – Brian M. Scott Oct 28 '13 at 8:36

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