Sequences and inequalities in $\mathbb{R}^n$ Let $x_k,y_k$ be sequences in $\mathbb{R}^n$ with $\lim x_k=a$, $\lim y_k=b$ and $$|x_k-b|<r<|y_k-a|$$ prove that $|a-b|=r.$
My attempt was:
We can write $|x_k-b+(a-a)|=|x_k-a+(a-b)|$ and $|y_k-a+(b-b)|=|y_k-b +(b-a)|$. My first question is: In $\mathbb{R}$ we can write $|u|-|v|\leq |u-v|$, can we do $$|x_k-a|-|b-a|\leq |x_k-a-(b-a)| ?$$
if we can, we have $$|x_k-a|-|b-a|<r<|y_k-b|+|b-a|$$
and $|x_k-a|<\epsilon,|y_k-b|<\epsilon.$ So we can write $$\epsilon-|b-a|<r<\epsilon +|b-a|$$ This was my last "step". I can't see how can i conclude that $|b-a|=r.$
 A: Proposition 1
The Euclidean norm defined on $\mathbb{R}^{n}$ is continuous.
Proof
Let $x_{k}$ be a sequence in $\mathbb{R}^{n}$ which converges to $x$. We must prove that $\|x_{k}\|$ converges to $\|x\|$.
Indeed, this is the case. Let $\varepsilon > 0$. Then there exists $n_{\varepsilon}\in\mathbb{N}$ s.t. for every $n\in\mathbb{N}$ one has that
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow \|x - x_{k}\| \leq \varepsilon
\end{align*}
Consequently, for the same $\varepsilon > 0$, there corresponds the same $n_{\varepsilon}\in\mathbb{N}$ s.t. for every $n\in\mathbb{N}$ we have that
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow |\|x\| - \|x_{k}\|| \leq \|x - x_{k}\| \leq\varepsilon
\end{align*}
and the result holds.
Proposition 2
Let $x_{k}$ be a sequence in $\mathbb{R}^{n}$ which converges to $x$. If $\|x - x_{k}\| < c$, then we have that
\begin{align*}
L = \lim_{k\to\infty}\|x - x_{k}\| \leq c
\end{align*}
Proof
Let us suppose otherwise that $L > c$. Let $\varepsilon = k(L-c)$, where $k\in(0,1)$.
Then there corresponds a $n_{\varepsilon}\in\mathbb{N}$ s.t.
\begin{align*}
n\geq n_{\varepsilon} & \Rightarrow |\|x - x_{k}\| - L| \leq \varepsilon\\\\
& \Rightarrow L - \varepsilon \leq \|x - x_{k}\| \leq c\\\\
& \Rightarrow L - c \leq k(L - c)\\\\
& \Rightarrow (1-k)(L - c) \leq 0\\\\
& \Rightarrow L \leq c
\end{align*}
which contradicts our original claim.
Therefore $L\leq c$. Similar reasoning applies when $\|x - x_{k}\| > c$.
Solution
Applying the continuity of the norm function and the property proposed above, we have
\begin{align*}
\|x_{k} - b\| < r < \|y_{k} - a\| & \Rightarrow \lim_{k\to\infty}\|x_{k} - b\| \leq r \leq \lim_{k\to\infty}\|y_{k} - a\|\\\\
& \Rightarrow \|\lim_{k\to\infty}(x_{k} - b)\| \leq r\leq \|\lim_{k\to\infty}(y_{k} - a)\|\\\\
& \Rightarrow \|a - b\| \leq r\leq \|b - a\|\\\\
& \Rightarrow r = \|a - b\|
\end{align*}
and we are done.
Hopefully this helps!
A: Your second last and last inequalities are true but not useful. Instead you want to use triangle inequality at the second last inequality as follows:
$|b-a|-\epsilon < |b-a|-|x_k-a|<r < |y_k-b|+|b-a|< |b-a|+\epsilon\implies |r-|b-a||<\epsilon\implies r = |b-a|$ since it is true for any $\epsilon > 0$.
