Exercise 1888Topology › Topology of ℝL1Added 3 Nov 2026

Continuity via open preimages

#continuity#open preimages#topological definition

Statement

Show that $f:\mathbb{R}\to\mathbb{R}$ is continuous iff $f^{-1}(U)$ is open for every open $U$ .

Hint

$(\Leftarrow)$ : apply to $U=]f(x_0)-\varepsilon,f(x_0)+\varepsilon[$ . $(\Rightarrow)$ : use the $\varepsilon$ - $\delta$ definition.

Show solutionHide solution

Solution

$\Leftarrow$ . Given $x_0$ and $\varepsilon>0$ , the ball $U=]f(x_0)-\varepsilon,f(x_0)+\varepsilon[$ is open, so $f^{-1}(U)$ is open. It contains $x_0$ , hence contains $B(x_0,\delta)\subseteq f^{-1}(U)$ . So $|f(x)-f(x_0)|<\varepsilon$ for $|x-x_0|<\delta$ .

$\Rightarrow$ . Given open $U$ and $x_0\in f^{-1}(U)$ , $f(x_0)\in U$ gives $\varepsilon>0$ with $B(f(x_0),\varepsilon)\subseteq U$ . Continuity provides $\delta>0$ with $B(x_0,\delta)\subseteq f^{-1}(U)$ .

Conclusion. Continuity is topological.

Continuity via open preimages

Statement

Solution

More problems in Topology of ℝ