Understanding a proof for existence of a pseudo-orbit trace for expanding circle map In Section 5.1 of Brin and Stuck's Introduction to Dynamical Systems they speak about infinite $\varepsilon$-orbits of $E_m=mx \mod 1$ on $S^1$, $m>1$. Consider a sequence $(x_n)_{n=0}^\infty$ such that $d(x_{n+1},E_mx_n)<\varepsilon$ for all $n\in \mathbb N_0$. They then go onto discuss how for each finite $n$ there is a point $y_n$ such that $d(E_m^jy_n,x_j)<\varepsilon$ for all $0\leq j\leq n$. I understand why such points exist. They then proceed to claim that $\lim_{n\to \infty}y_n$ exists. I am not entirely sure why.
Let me elucidate on my confusion. As for each $n$ we can construct a $y_n$ with the required properties, we are justified in considering the sequence $(y_n)$. Now, as $S^1$ is a complete, compact metric space it follows that $(y_n)$ has some convergent subsequence $(y_{n_k}) \to y$. Now for any $j\in \mathbb N$ we know that $k>j$ implies that $d(E^j_my_{n_k},x_j)<\varepsilon$. Furthermore $d(E^j_my,x_j)\leq d(E^j_my, E^j_my_{n_k})+d(E^j_my_{n_k},x_j)$. Taking the limit as $k$ goes to $\infty$ on both sides thus gives $d(E_m^j y,x_j)\leq \varepsilon$. As $j$ was arbitrary it follows that the orbit of $y$ is indeed a $\varepsilon$ trace of the pseudo-orbit $(x_n)$. However, I cannot convince myself that this $y$ is actually the limit of $(y_n)$ in the traditional sense. I.e why can't there be another subsequence $(y_{n_j})$ that doesn't converge to $y$? If someone could explain to me why this is the case it would be much appreciated.
 A: Lemma: In a metric space $(X,d)$, if every subsequence of a sequence $(y_n)$ has a further subsequence converging to some $y\in X$, then $(y_n)$ converges to $y$.
Proof: Suppose, by way of contradiction, that $(y_n)$ does not converge to $y$. Then there exists a $\varepsilon>0$ such that for every $n\in \mathbb N$ there exists a $k>n$ such that $d(y_k,y)\geq \varepsilon$. Now we can choose the subsequence $(y_{n_j})$ of $(y_n)$, where for each $j\in \mathbb N$, $n_j$ is the smallest number bigger than $j$ such that $d(y_{n_j},y)\geq\varepsilon$. The well-ordering of the naturals and our assumption ensures that this subsequence exists. By the very definition of a subsequence, we know for any subsequence $(y_{n_{j_k}})$ of $(y_{n_j})$ that $d(y_{n_{j_k}},y)\geq \varepsilon$ for all $k\in \mathbb N$. Hence no subsequence of $(y_{n_j})$ can converge to $y$, contradicting the fact that every subsequence has a subsequence converging to $y$.
Returning to the problem, as $S^1$ is compact, we know that each subsequence $y_{n_j}$ has a convergent subsequence $y_{n_{j_l}}$. As John B pointed out in the comments, the expansive nature of $E_m$ means that each $y_{n_{j_l}}$ must converge to $y$. The above lemma then gives the result.
