How does the given proof is a spoof? [Completeness is a topological property of metric spaces]

Question

Let $f:(X,d)\to (Y,e)$ homeomorphism. So, $f^{-1}$ is continuous. So $f^{-1}$ is continuous at $y \in Y$, For any $\epsilon \in \mathbb R^+$, $\exists \delta\in \mathbb R^+$:$e(x,y)<\delta \implies d(f^{-1}(x),f^{-1}(y))<\epsilon. $ Let $\{y_n\}$ be a cauchy sequence in $(Y,e)$. For the same $\delta$, There is a $N\in \mathbb N$: $\forall m,n\ge N\implies e(y_n,y_m)<\delta$. Define a sequence $\{x_n=f^{-1}(y_n)\}$. $\forall m,n\ge N \implies d(f^{-1}(y_n),f^{-1}(y_m))<\epsilon.$ Hence $\{x_n\}$ converges in $(X,d)$. So $f(x_n)$ converges to $f(x)$. Since $f$ is continuous. Hence $\{y_n\}$ converges to $f(x)$. How do my argument go wrong?

Answer 1

In your argument, you are using the uniform continuity criteria. In general, given $\epsilon \gt 0$, same $\delta $ will not work for every point.