Here is a lemma and a proof given to me in class.

Lemma If $M$ is a smooth manifold, $K\subseteq M$ a compact subset, $A\subset M$ an open set containing $K$< then there exists a compact-support smooth function $\chi\in\mathcal{C}^\infty_0(M)$ such that $\mathrm{supp}\,\chi\subseteq A$ and $\chi\equiv 1$ on $K$.

Proof Let $B:=M\smallsetminus K$. $B$ is then open, so $\{A,B\}$ is a cover of $M$. It is also finite, thus locally finite, thus we can find a partition of unity $\{\phi_A,\phi_B\}$ such that $\phi_A+\phi_B\equiv1$ on all of $M$, $\mathrm{supp}\,\phi_A\subseteq A$ and $\mathrm{supp}\,\phi_B\subseteq B$. If we set $\chi:=\phi_A$, we have $\mathrm{supp}\,\chi\subseteq A$, and $\chi|_K\equiv1$, since $K\subseteq M\smallsetminus B$ whence $\phi_B|_K\equiv0,\chi|_K=\phi_A|_K=(1-\phi_B)|_K\equiv1-0=1$. We only need to make $\mathrm{supp}\,\chi$ compact, and it is here that I have the problematic passage. We are told to find an open set $A'$ such that $K\subseteq A'\subseteq A$ and $\overline{A'}$ is compact. Assuming this can be found, then if we construct the partition of unity from $\{\phi_{A'},\phi_B\}$ and set $\chi:=\phi_{A'}$, $\mathrm{supp}\,\chi$ will be contained in $\overline{A'}$, which is compact, and as the support is the closure of the set of non-zeros it is closed, and being closed in a compact set means compactness, hence the proof is concluded. $\hspace{2cm}\square$

How do I prove there exists such an $A'$? I mean, I can easily find an open set $A'$ satisfying the inclusion, but how do I find one that is relatively compact? Is it true that, given $M$ a manifold, $K$ a compact set in $M$ and $A$ an open set containing $K$, there is always a relatively compact open set in between the two?

  • $\begingroup$ By the way, does this lemma bear the name of a mathematician? Because I missed the lesson it was proved in, and a classmate of mine's notes call it «Lemma di Biase» (Biase's lemma), or so I seem to read, so do you know if it has a name? $\endgroup$ – MickG Apr 3 '15 at 11:00
  • $\begingroup$ Why the downvote?? $\endgroup$ – MickG Apr 3 '15 at 12:37

A topological space is called locally compact if every point has a compact neighbourhood. The fact that you want to use is the following.

Suppose $X$ is locally compact and Hausdorff, and $K \subset U \subset X$ where $K$ is compact and $U$ is open. Then there is a relatively compact set $V$ such that $K \subset V \subset \overline{V} \subset U$.

Note, $V$ relatively compact means that $\overline{V}$ is compact. The proof of this result is straightforward and can be found in Section $4.5$ of Folland's Real Analysis: Modern Techniques and Their Applications (second edition) for example.

As well as being Hausdorff, every topological manifold is locally Euclidean and therefore locally compact, so the above result applies to manifolds.

I would call the Lemma a smooth version of the Urysohn Lemma.

  • $\begingroup$ Um, "compact" and "neighborhood" do not go too well together in my mind, since "compact" implies closed, and "neighborhood" means "open set" to me. Perhaps I should review my definition of neighborhood. Would you expand on that? $\endgroup$ – MickG Apr 3 '15 at 13:56
  • $\begingroup$ Here a neighbourhood of $x$ is a set which contains an open set which contains $x$. $\endgroup$ – Michael Albanese Apr 3 '15 at 14:04
  • 1
    $\begingroup$ Oh, and «every topological manifold is locally Euclidean and therefore locally compact», i.e. every point has a neighborhood homeomorphic to an open set in $\mathbb{R}^n$, which is locally compact, so I can find a compact neighborhood of the image of that point which can be mapped back via the homeomorphism to a compact neighborhood of the starting point of the manifold. This was just for myself. $\endgroup$ – MickG Apr 3 '15 at 14:08
  • $\begingroup$ @MickG: I'm pretty sure the link you posted is to an illegally pirated copy of Folland's book. $\endgroup$ – Jack Lee Apr 3 '15 at 14:57
  • $\begingroup$ Let's say I delete that comment and post a couple of screenshots with the relevant parts. 1 and 2. That's all that's needed. $\endgroup$ – MickG Apr 3 '15 at 15:22

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