Open Unit Interval is Proper Subset of Closed Unit Interval

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Theorem

The open unit interval:

$I_o = \left({0 \,.\,.\, 1}\right)$

is a proper subset of the closed unit interval:

$I_c = \left[{0 \,.\,.\, 1}\right]$


Proof

Let $x \in I_o$.

Then by definition:

$0 < x < 1$

and so:

$0 \le x \le 1$

and so:

$x \in I_c$.

Thus:

$I_o \subseteq I_c$

Consider:

$0 \in I_c$

by definition of closed interval.

But it is not the case that $0 < 0$.

So $0 \notin I_o$ and so $I_c \not \subseteq I_o$.

Hence the result by definition of proper subset.

$\blacksquare$


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