Given two metric spaces \((X,d_X)\) and \((Y,d_Y)\). A function \(f\colon X\to Y\) is continuous (in the metric sense) if and only if it is continuous (in the topological sense).

Proof

Let \(U\in Y\) be open and \(x\in f^{-1}(U)\). Then there is \(B_\varepsilon(f(x))\in U\) and since \(f\) is metrically continuous there is a \(\delta>0\) such that \(B_\delta(x)\subseteq f^{-1}(B_\varepsilon(f(x))\subseteq f^{-1}(U)\). Therefore \(f^{-1}(U)\) is open.

Conversely, let \(\varepsilon>0\). Since \(B_\varepsilon(f(x))\) is open the inverse image is also open. That means there is \(\delta>0\) such that \(B_\delta(x)\subseteq f^{-1}(B_\varepsilon(f(x))\). In other words \(f\) is metrically continuous.

Remark
According to [1] this observation motivated the definition of topological spaces.

References Link to heading

  1. J. Lee, Introduction to Topological Manifolds. New York, NY: Springer New York, 2011. doi:10.1007/978-1-4419-7940-7