Let $(E, d)$ be a metric space and $A$ a non-empty subset of $E$. I'm trying to give an alternative proof to this result, i.e.,

Theorem: If $A$ is open in $E$, then there is a sequence $(f_n)$ of real-valued bounded continuous functions on $E$ such that $f_n \to f$ pointwise.

Could you have a check on my attempt?

Proof: Let $B := E\setminus A$ and $f_n (x) := \sqrt[n]{\min \{d(x, B), 1\}}$ where $d(x, B) := \inf_{b \in B} d(x, a)$. Notice that $|d(x, B) - d(y, B)| \le d(x, y)$, so $d(\cdot, B)$ is continuous. As such, $\min \{d(\cdot, B), 1\}$ is bounded continuous. Let's prove that $f_n \to f$ pointwise.

  • Let $x \in A$. Then $x \notin B$. Because $B$ is closed, $d(x, B)>0$. Then $\alpha := \min \{d(x, B), 1\} \in (0, 1]$. Notice that $\ln \alpha \le 0$. It follows that $$ f_n (x) = \alpha^{1/n} = e^{\frac{\ln \alpha}{n}} \nearrow 1 = 1_A (x) \quad \text{as} \quad n \to \infty. $$

  • Let $x \notin A$. Then $x \in B$ and thus $f_n (x) =0 =1_A (x)$. This completes the proof.

  • 1
    $\begingroup$ Looks ok to me. $\endgroup$
    – Didier
    Oct 24, 2022 at 17:09


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