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.