Discussion
Let \(d_{1}\) and \(d_{2}\) denote the usual metric and the discrete metric on \(\mathbb{R}\), respectively. Let \(f : (\mathbb{R}, d_{1}) \to (\mathbb{R}, d_{2})\) be defined by \(f(x) = x,\ x \in \mathbb{R}\). Then
Answer: B
\(f^{-1}\) is continuous but \(f\) is NOT continuous.
In the discrete metric every set is open, in particular every singleton \(\{a\}\). Its preimage under \(f\) is the singleton \(\{a\}\) in the usual metric, which is not open. So \(f\) is not continuous.
\(f^{-1} : (\mathbb{R}, d_{2}) \to (\mathbb{R}, d_{1})\) is the identity from a discrete space. Every function on a discrete space is continuous, because every preimage is automatically open.
That rules out the options saying \(f\) is continuous, and the one saying neither map is continuous.
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