Let $S\subset U$. What does it mean to say that $S$ is relatively closed in $U$? Also $U\subset\mathbb{R}^{n}$ is open and bounded, but I don't know if that's essential.

Here follows an example from where I found the term, a sort of "use it in a sentence", if you like.

Theorem 4 from chapter 2 of Partial Differential Equations by Lawrence C. Evans ($U\subset\mathbb{R}^{n}$ is open and bounded):

Strong maximum principle. Suppose $u\in C^{2}\left(U\right)\cap C\left(\bar{U}\right)$ is harmonic within $U$.

(1) Then

$$\max_{\bar{U}}u=\max_{\partial U}u$$

(2) Furthermore, if $U$ is connected and there exists a point $x_{0}\in U$ such that



$$u\text{ is constant within }U.$$

Proof. Suppose there eixsts a point $x_{0}\in U$ with $u\left(x_{0}\right)=M:=\max_{\bar{U}}u$. Then for $0<r<\text{dist}\left(x_{0},\partial U\right)$, the mean-value property asserts

$$M=u\left(x_{0}\right)=\frac{1}{V}\int_{B\left(x_{0},r\right)}udy\leq M.$$

As equality holds only if $u\equiv M$ within $B\left(x_{0},r\right)$, we see that $u\left(y\right)=M$ for all $y\in B\left(x,r\right)$. Hence the set $\left\{ x\in U|u\left(x\right)=M\right\}$ is both open and relatively closed in $U$, and thus equals $U$ if $U$ is connected. This proves assertion (2), from which (1) follows. $\blacksquare$

Thus, I need to know: What does it mean to say that $S\subset U$ is relatively closed in $U$? And, why if $U$ is connected it follows that $S=U$?


  • 5
    $\begingroup$ It means it is closed in the topology of your subset. A relatively closed set of $U$ is the intersection of a closed set and $U$. e.g. $[.5,1)$ is relatively closed in $(0,1)$. $\endgroup$ – ShawnD Feb 29 '12 at 15:27

This should mean that $S$ is a closed subset of the topological space $U$, where the topology on $U$ is the subspace topology it gains as a subset of $\mathbf R^n$. Explicitly, this means that there is a closed subset $\widetilde S$ of $\mathbf R^n$ such that $S = U \cap \widetilde S$. As Shawn notes in the comments, a good example is the relatively closed subset $[1/2, 1)$ of the interval $(0, 1)$.

Note that $U - S$ is also open and closed in $U$. If $U - S$ were nonempty, then $U = S \cup (U - S)$ would contradict the connectedness of $U$.

  • 6
    $\begingroup$ Maybe it's helpful to point out that we don't need to say "relatively open" because a relatively open subset of an open subset is already open in the ambient space. $\endgroup$ – Dylan Moreland Feb 29 '12 at 15:45

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