Limit of Tail of Decreasing Sequence of Sets

Theorem

Let $X$ be a set.

Let $\sequence {E_n}_{n \mathop \in \N}$ be a decreasing sequence of subsets of $X$ such that:

$E_n \downarrow E$

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

Then for each $m \in \N$ we have:

$E_{n + m} \downarrow E$

Proof

Let $m \in \N$.

From Tail of Decreasing Sequence of Sets is Decreasing, we have:

$\sequence {E_{n + m} }_{n \mathop \in \N}$ is a decreasing sequence of sets.

Since:

$E_n \downarrow E$

we have:

$\ds \bigcap_{n \mathop = 1}^\infty E_n = E$

We show that:

$\ds \bigcap_{n \mathop = 1}^\infty E_{n + m} = E$

That is:

$\ds \bigcap_{n \mathop = m + 1}^\infty E_n = E$

Clearly we have:

$\ds \bigcap_{n \mathop = 1}^\infty E_n \subseteq \bigcap_{n \mathop = m + 1}^\infty E_n$

from Intersection is Decreasing.

Now let:

$\ds x \in \bigcap_{n \mathop = m + 1}^\infty E_n$

Then:

$x \in E_n$ for all $n \in \N$ for all $n \ge m + 1$

and in particular:

$x \in E_{m + 1}$

Since $\sequence {E_n}_{n \mathop \in \N}$ is a decreasing sequence of sets, we have:

$E_{m + 1} \subseteq E_n$ for all $n \in \N$ with $n < m + 1$.

So, we actually have:

$x \in E_n$ for all $n \in \N$.

So, from the definition of set intersection, we have:

$\ds x \in \bigcap_{n \mathop = 1}^\infty E_n$

So, from the definition of set inclusion, we have:

$\ds \bigcap_{n \mathop = m + 1}^\infty E_n \subseteq \bigcap_{n \mathop = 1}^\infty E_n$

So we have:

$\ds \bigcap_{n \mathop = 1}^\infty E_{n + m} = E$

$\blacksquare$