# Singular support of a tempered distribution is compact?

I am reading Introduction to the Theory of Distributions by Friedlander and Joshi. As definition 8.6.1, they define the singular support of a tempered distribution $u$ to be the complement of {$x$: $u$ is $C^{\infty}$ on some neighborhood of $x$}.

From this definition I cannot see why the singular support of a tempered distribution need to be compact. But in proof for Lemma 8.6.1, they begin by choosing a test function $\phi$ such that $\phi\equiv 1$ on the singular support of a given tempered distribution.

Are they assuming that the singular support of a tempered distribution is compact? How can they do this?

Thanks!

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They can't. And it is probably a mistake in the book. The analogous statement in Hormander's Analysis of Linear Partial Differential Operators volume 1 is Theorem 4.2.5, where instead of $\mathscr{S}'$ the statement is given in terms of $\mathscr{E}'$, the space of distributions with compact support.
BTW, if you look at two pages on where the Lemma 8.6.1 is used, it is applied to $E * P\psi u$ where $E\in \mathscr{S}'$, $P$ is a linear partial differential operator, $u\in \mathscr{D}'$, but $\psi$ is a smooth cutoff function. So necessarily $P\psi u\in \mathscr{E}'$.