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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

$$u\left(x_{0}\right)=\max_{\bar{U}}u$$

then

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

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 $S=U$?

Thanks.

a06e
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    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 $(0,1)$. – ShawnD Feb 29 '12 at 15:27

1 Answers1

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"$S$ is relatively closed in $U$" means 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 $\mathbf R^n$.

Clearly, $S$ is relatively closed in $U$ if and only if there is a closed subset $\widetilde S$ of $\mathbf R^n$ such that $S = \widetilde S\cap U$.

As Shawn notes in the comments, a good example is the relatively closed subset $[1/2, 1)$ of the interval $(0, 1)$, because $[1/2,1)=[1/2,2]\cap(0,1)$.

However, relatively open does not mean that but something completely different.

user3810316
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Dylan Moreland
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    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. – Dylan Moreland Feb 29 '12 at 15:45
  • How $U - S$ is closed in $U$? – Hermis14 Oct 18 '22 at 13:04
  • $U\setminus S$ need not be closed in $U$; e.g., $[0,1]^c$ is not closed in $\mathbb R$. I removed the erroneous paragraph. I added that "relatively open" is not defined analogously; i.e., it does not mean "complement of relatively closed". It means "equal to its relative interior", which is defined with respect to its affine hull. See the link. Thus, Dylan's comment assumes the wrong definition and is, hence, incorrect (though that is what I'd guessed too before I knew). – user3810316 Nov 29 '24 at 11:56