What does the expression 'a closed subset of a set' imply? This question is about the terms: a closed or open subset of some set.
If we say that $A$ is an open subset of the topological space $(X, \tau)$, then, it is apparent that $A$ is open in $X$, that is, $A \in \tau$.
On the other hand, consider any subset $B $ of $ X$, and say $C$ is a closed subset of $B$. Now, I guess there are two possible interpretations:

*

*$ C $ is a subset of $B$ that is closed in $X$;

*$C$ is a subset of $B$, and $ C $ is relatively closed in $B$, i.e., $C$ is closed in the subspace $B$ of $X$.

In the second case, $C$ is not necessarily closed in $X$.
I came up with this question when reading the article The Theorems of Bony and Brezis on Flow-Invariant Sets. On the first page, the author says $F$ is a closed subset of $\Omega$, where $\Omega$ is a subset of the Euclidean space $\mathbb{R}^n$. If I accept the first interpretation, I would say the condition on $F$ is too restrictive, especially when $\Omega$ is open in $\mathbb{R}^n$, which is quite likely for dynamical systems.
In the original paper,

Throughout this note $\Omega$ is a domain in real Euclidean space $E_n$, $X(x)$ is a function on $\Omega$ to $E_n$, and $F$ is a closed subset of $\Omega$.

But the source from which I encounter this confusion is not restricted to the referenced paper. I want to know what is the usually accepted interpretation in the literature.
If there is any possibility that it depends on the context, I would like to have comments about which one is correct from someone familiar with the theory about the invariance of a set with respect to dynamical systems.
Update: It turned out from a paper the author cited that the second one was what he meant. Refer to On a characterization of flow-invariant sets, where the author states that

Let $E$ be a finite-dimensional Euclidean space. Let $\Omega$ be an open set in $E$ and let $F \subseteq \Omega$ be relatively closed in $\Omega$.

After all, this is satisfying because the second interpretation looked much more reasonable to me. But I am still not sure what is naturally accepted.
 A: Let $X$ be a topological space and let $C\subset B\subset X$ be subsets.

*

*If $C$ is closed in $X$ then $C=X\setminus C_o$ for some open $C_o\in\tau_X$. The open sets in the subspace topology $\tau_B$ of $B$ are of the form $S_B=S_o\cap B$ for some $S_o\in\tau_X$. Since $C\subset B$ we have
$$
C=
X\setminus C_o=
(X\setminus C_o)\cap B=
(X\cap B)\setminus (C_o\cap B)=
B\setminus (C_o\cap B),
$$
i.e. $C$ is a complement of an open set in the subspace topology, so $C\subset B$ is closed for the subspace topology on $B$.


*Let $C\subset B$ be closed in $B$ for the induced subspace topology, i.e. $C=B\setminus(C_o\cap B)$ for some open $C_o\in\tau_X$. If $B$ is closed then $B=X\setminus B_o$ for some $B_o\in\tau_X$. Thus:
$$
C=
B\setminus(C_o\cap B)=
B\setminus C_o=
(X\setminus B_o)\setminus C_o=
X\setminus (B_o \cup C_o).
$$
Since $B_o \cup C_o\in\tau_X$ this shows, that in this case indeed $C$ is closed for the topology on $X$.
As you already stated, if $B$ is not closed, there is nothing to say in general. $C$ may or may not be closed for the topology on $X$.
From this you see: in any of the two cases $C$ is closed for the subspace topology on $B$. The reverse is however not given. In particular for your application when $F$ is a "domain" usually one has $F$ being open in $X$. This does not exclude closedness per se, but in many cases open domains are not closed.
