Convergent Sequence in Metric Space is Bounded

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Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $\sequence {x_n}$ be a sequence in $M$ which is convergent, and so $x_n \to l$ as $n \to \infty$.


Then $\sequence {x_n}$ is bounded.


Convergent Sequence in Normed Division Ring is Bounded

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.

Let $\sequence {x_n}$ be a sequence in $R$.

Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limit:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$


Then $\sequence {x_n}$ is bounded.


Convergent Real Sequence is Bounded

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

Let $l \in A$ such that $\displaystyle \lim_{n \mathop \to \infty} x_n = l$.


Then $\left \langle {x_n} \right \rangle$ is bounded.


Proof

Let $M = \struct {A, d}$ be a metric space.

Let $\sequence {x_n}$ be a sequence in $M$ which is convergent, and so $x_n \to l$ as $n \to \infty$.

From the definition, in order to prove boundedness, all we need to do is find $K \in \R$ such that $\forall n \in \N: \map d {x_n, l} \le K$.

Since $\sequence {x_n}$ converges, it is true that:

$\forall \epsilon > 0: \exists N: n > N \implies \map d {x_n, l} < \epsilon$

In particular, this is true when $\epsilon = 1$, for example.

That is:

$\exists N_1: \forall n > N_1: \map d {x_n, l} < 1$

So now we set:

$K = \max \set {\map d {x_1, l}, \map d {x_2, l}, \ldots, \map d {x_{N_1}, l}, 1}$

The result follows.

$\blacksquare$


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