Metric Space (Elementary Analysis)

Question

Let $X \subseteq \mathbb{R}^{n}$ be given by

$$ X = \left\{ (b_{1}, \ldots, b_{n}) \in \mathbb{R}^{n} \mid \sum\limits_{i=1}^{n} \frac{b_{i}}{i} = 0 \right\}$$

Then prove that $X$ is closed in $\mathbb{R}^{n}$.

Answer 1

HINT: Convergence in $\mathbb{R}^{n}$ is equivalent to coordinatewise convergence.

Answer 2

There are many ways to do this :

1. Take a sequence $b^k = (b_1^k, b_2^k, \ldots, b_n^k) \in X$ such that $b^k \to b = (b_1, b_2, \ldots, b_n)$. Then $b_i^k \to b_i$ for each $i$, and hence $$ \sum_{i=1}^n \frac{b_i^k}{i} \to \sum_{i=1}^n \frac{b_i}{i} $$ and hence the RHS is zero.

2. Consider the function $f:\mathbb{R}^n \to \mathbb{R}$ by $$ f(b_1, b_2,\ldots, b_n) = \sum_{i=1}^n \frac{b_i}{i} $$ Then $f$ is continuous (because of part 1), and $$ X = f^{-1}(\{0\}) $$

Take your pick.