Limit of Subsequence of Bounded Sequence

Theorem
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

Let $\left \langle {x_n} \right \rangle$ be bounded.

Let $b \in \R$ be a real number.

Suppose that $\forall N: \exists n > N: x_n \ge b$.

Then $\left \langle {x_n} \right \rangle$ has a subsequence which converges to a limit $l \ge b$.

Proof
Let us pick $N \in \N$.

Then $\exists n_1 > N: x_{n_1} \ge b$.

Again, $\exists n_2 > n_1: x_{n_2} \ge b$.

And so on: for each $n_k$ we find, $\exists n_{k+1} > n_k: x_{n_{k+1}} \ge b$.

In this way we can build a subsequence of $\left \langle {x_n} \right \rangle$ each of whose terms are $b$ or bigger.

By the Bolzano-Weierstrass Theorem, this subsequence itself contains a subsequence $\left \langle {x_{n_r}} \right \rangle$ which is convergent.

Now, suppose $x_{n_r} \to l$ as $r \to \infty$.

Since $x_{n_r} \ge b$ it follows from Lower and Upper Bounds for Sequences that $l \ge b$.