Limit of Decreasing Sequence of Left Half-Open Intervals with Lower Bound Converging to Upper Bound

Theorem

Let $a, b \in \R$ have $a < b$.

Let $\sequence {a_n}_{n \mathop \in \N}$ be an increasing sequence with $a_n \to b$.

Then we have:

$\ds \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b = \set b$

That is:

$\hointl {a_n} b \downarrow \set b$

where $\downarrow$ denotes the limit of decreasing sequence of sets.

Proof

Clearly:

$\ds b \in \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b$

so that:

$\ds \set b \subseteq \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b$

Now let:

$\ds x \in \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b$

Then:

$a_n \le x \le b$

for each $n \in \N$.

From Limits Preserve Inequalities, we then have:

$b \le x \le b$

so that:

$x = b$

So:

$\ds \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b = \set b$

as required.

$\blacksquare$