I know it's a well-known result, but I have not found any clear formalization, and I need a clear formalization. So I want to know if you agree with this formalization, and this proof.

Thank you for any replies.

Let $K \subset \Omega \subset \mathbb{R}^n$ with $K$ compact and $\Omega$ open. Then there exists a regular Urysohn function $\psi \in \mathcal{D}(\Omega)$, i.e. $0 \leq \psi(x) \leq 1$ $\forall x \in \Omega$ e $\psi(x)=1$ on a neighborhood of $K$, and this fact is denoted by $ K \prec \psi \prec \Omega$. My proof is the following:
Since $K$ is closed, if $\widetilde{x} \in \mathbb{R}^n \setminus \Omega \subset \mathbb{R}^n \setminus K$, for the properties of the distance we have $0 < \delta /2 < \mathrm{dist}(K,\Omega^c)=\widetilde{\delta}$. We assume that $\delta/2 < < \widetilde{\delta}$.

Let $\varphi \in \mathcal{D}(\Omega)$ with $\mathrm{supp}(\varphi) \subseteq \overline{B}(0,\delta /2)$. without considering the translations, we define: \begin{align*} K(\delta /2 ):=K + \overline{B}(0,\delta /2):=\lbrace y \in \mathbb{R}^n : \mathrm{dist}(y,K) \leq \delta \rbrace \end{align*} In other words, $\lbrace K(\delta /2) \rbrace_{\delta > 0}$ it's an increasing sequence of compact in $\Omega$ for $\delta \rightarrow \widetilde{\delta}$ with $K \subset K(\delta /2) \subset \Omega$. Define the function: \begin{align*} \psi(x):=( \chi_{K(\delta /2)} \ast \varphi)(x)=\int_{\mathbb{R}^n} \chi_{K(\delta /2)}(x-y) \varphi(y) dy \end{align*} Then $\psi(x)$ satisfies the required properties. In fact, we know that $\psi \in \mathcal{E}(\mathbb{R}^n)$ and also \begin{align*} \mathrm{supp}(\psi) &= \mathrm{supp}(\chi_{K(\delta /2)} \ast \varphi) \\ &\subset \mathrm{supp}(\chi_{K(\delta /2)})+ \mathrm{supp}(\varphi) \\ &\subset K(\delta /2) + \overline{B}(0,\delta /2) \\ &= K+ \overline{B}(0,\delta /2) + \overline{B}(0,\delta /2) \\ &= K+ \overline{B}(0,\delta) \\ &=K(\delta) \end{align*} Consequently, $\psi \in \mathcal{D}(\Omega)$, and $0 \leq \psi(x) \leq 1$ $\forall x \in \Omega$. and $\psi(x)=1$ $\forall x \in K$.


1 Answer 1


The usual statement of Urysohn's lemma is that we can find a function $\psi$ with $\operatorname{supp}\psi\subseteq\Omega$ and $\psi=1$ in $K$. This is different from what you stated. Other than that, the proof of the theorem is correct. Simply use $K$ instead. I would also change $\delta/2$ by $\delta/3$ in the arguments, just to be on the safe side (since then we can take some closures, etc...), but this is not really necessary.

  • $\begingroup$ I used $\delta /2$ only to find that $\mathrm{supp}(\psi) \subset K(\delta)$. In the end, everything depends on the choice $\delta /2 << \widetilde{\delta}$. $\endgroup$
    – user288972
    Feb 16, 2016 at 23:02
  • $\begingroup$ Anyway, you say $\psi=1$ in $K$, but $K$ is compact, so is a locally compact space, and then the lemma is also true in a neighborhood (compact) of $K$. The sense is this. $\endgroup$
    – user288972
    Feb 25, 2016 at 20:04
  • 1
    $\begingroup$ @JohnMartin I was just pointing out the usual statement of Urysohn's lemma. Sure, we can take $L$ compact with $K\subseteq\operatorname{int}L\subseteq L\subseteq\Omega$, apply this ``usual'' version to $L$ and obtain $\psi=1$ in a neighbourhood of $K$ (namely, $\operatorname{int}L$). However in your proof you just get $\psi=1$ in $K$, but not in a neighbourhood of $K$. Also, you didn't mention, in your statement, that $\psi$ should be zero outside $\Omega$. All that said, your proof is correct, the only problem is with the theorem's statement. $\endgroup$ Feb 26, 2016 at 21:14
  • $\begingroup$ Yes, I agree with these considerations $\endgroup$
    – user288972
    Feb 26, 2016 at 21:42

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