Convergent Subsequence in Closed Interval

From ProofWiki
Jump to: navigation, search

Theorem

Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.


Then every sequence of points of $\left[{a \,.\,.\, b}\right]$ contains a subsequence which converges to a point in $\left[{a \,.\,.\, b}\right]$.


Proof

Let $\left \langle {x_n} \right \rangle$ be a sequence in $\left[{a \,.\,.\, b}\right]$.

Since $\left[{a \,.\,.\, b}\right]$ is bounded in $\R$, it follows that $\left \langle {x_n} \right \rangle$ is a bounded sequence.

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


Suppose $x_{n_r} \to l$ as $n \to \infty$.

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


So $\left \langle {x_{n_r}} \right \rangle$ converges to a point in $\left[{a \,.\,.\, b}\right]$.

$\blacksquare$


Sources