Let $(\Omega, d)$ be a metric space. A sequence $\left(x_n\right)_{n \in \mathbb{N}}$ in $\Omega$ converges to an element $x \in \Omega$ if $d\left(x_n, x\right) \rightarrow 0$ as $n \rightarrow \infty$, i.e., $$ \forall \epsilon>0 \exists N \in \mathbb{N} \text { such that } \forall n \geq N: d\left(x, x_n\right)<\epsilon . $$
Prove that the sequence $x_n=1 / n, n \in \mathbb{N}$ does not converge if $\Omega=(0,1], d(x, y)=|x-y|$.
Below I write my attempt. It seems right to me but I have the impression that this exercise could be solved much more easily. What do you think about it?
To prove that the sequence $x_n = \frac{1}{n}, n \in \mathbb{N}$ does not converge in the metric space $(\Omega, d)$, where $\Omega = (0, 1]$ and $d(x, y) = |x - y|$, we can use a proof by contradiction.
Let's assume, for the sake of contradiction, that this sequence converges to a certain point $x$ in $(0, 1]$. This would imply that for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for every $n \geq N$, $d(x, x_n) < \epsilon$, i.e., $|x - \frac{1}{n}| < \epsilon$. But by choosing $\epsilon = \frac{x}{2}$, we obtain that there exists an $N$ such that for every $n \geq N$, $|x - \frac{1}{n}| < \frac{x}{2}$.
Now, let's consider an $n$ such that $n > \frac{2}{x}$. In this case, we have: $$ |x - \frac{1}{n}| < \frac{x}{2} \quad \Rightarrow \quad |x - \frac{1}{n}| - x < -\frac{x}{2} \quad \Rightarrow \quad -\frac{1}{n} < -\frac{x}{2} \quad \Rightarrow \quad \frac{1}{n} > \frac{x}{2} \quad \Rightarrow \quad n < \frac{2}{x} $$ However, this inequality contradicts the initial choice of $n$ ($n > \frac{2}{x}$). Therefore, we have reached a contradiction. This means that our initial assumption, i.e., the sequence $x_n = \frac{1}{n}$ converges to a certain point $x$ in $(0, 1]$, is false.
Hence, the sequence $x_n = \frac{1}{n}$ does not converge in the metric space $(\Omega, d)$ with $\Omega = (0, 1]$ and $d(x, y) = |x - y|$.
