Is the identity id:ℝ→ℝ, from the usual topology to the discrete one, continuous?

No: \0\ is open in the discrete topology, but id^-1(\0\)=\0\ is not open in the usual one

Yes, because the discrete topology has more open sets

No, because id is not injective

Exercise · ZW-CONTINUOUS-MAP-TOPOLOGICAL-010 · ★★★ Advanced

Is the identity \(\mathrm{id}:\mathbb{R}\to\mathbb{R}\), from the usual topology to the discrete one, continuous?

Choose one option.

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C. No: \(\{0\}\) is open in the discrete topology, but \(\mathrm{id}^{-1}(\{0\})=\{0\}\) is not open in the usual one

Explanation

To refute continuity it suffices to find an open set of the target whose preimage is not open. Here \(\{0\}\) does. Having more open sets in the target makes continuity harder, not easier; and the identity is injective.

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Review the concept: Continuous function between topological spaces

A function \(f:X\to Y\) between topological spaces \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) is continuous if the preimage of every open set \(V\) of \(Y\) is an open set of \(X\).

\[V\in\mathcal{T}_Y\ \Longrightarrow\ f^{-1}(V)\in\mathcal{T}_X\]

Is the identity \(\mathrm{id}:\mathbb{R}\to\mathbb{R}\), from the usual topology to the discrete one, continuous?

Explanation

Review the concept: Continuous function between topological spaces

More practice problems: continuous function between topological spaces