Question

Please show that any convergent sequence in a metric space is bounded, then find a bounded sequence of real numbers that is not convergent.

Answer 1

Question 1. Please show that any convergent sequence in a metric space is bounded, then find a bounded sequence of real numbers that is not convergent.

Solution. Let $x_{n} \to a$ as $n \to \infty$ in a metric space $(X, \rho)$. By definition of convergence for any $\varepsilon > 0$ there is $N \in \mathbb{N}$ such that $\rho(x_{n}, a) < \varepsilon$ for all $n > N$. Take $\varepsilon = 1$ and find the corresponding $N \in \mathbb{N}$. Set $M$ to be $\max \{\rho(x_{1}, a), \ldots, \rho(x_{N}, a), \varepsilon\}$. Then for all $n \geq 1$ we have $\rho(x_{n}, a) \leq M$. Thus, the sequence $\{x_{n}\}_{n \in \mathbb{N}}$ is contained in the closed ball of radius $M$ with the center in $a$. So, $\{x_{n}\}_{n \in \mathbb{N}}$ is bounded.

For an example of a bounded real sequence which is not convergent consider $x_{n} = (-1)^{n}$. We have $|x_{n}| = 1$ for all $n$, but there are two subsequences $x_{2n} = 1$ and $x_{2n - 1} = -1$, $n \geq 1$, which are obviously convergent and have different limits. This explains why $x_{n}$ cannot be convergent.

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