3
$\begingroup$

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:

  1. $ C $ is a subset of $B$ that is closed in $X$;
  2. $C$ 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.

$\endgroup$
1
  • $\begingroup$ I claim the answer is definitely your second interpretation. The sentence "A is closed in B" usually means that A is a subset of the topological space B. It is common to treat subsets of topological spaces as synonymous with the subspace they define. So "B" would be understood to be the subspace of X defined by (the actual subset) B, and A would be taken to be closed relative to it. This is just my gut feeling - I don't have references to back it up (or I'd post an answer). But I'd put money on it. $\endgroup$ Aug 5, 2022 at 8:27

1 Answer 1

0
$\begingroup$

Let $X$ be a topological space and let $C\subset B\subset X$ be subsets.

  1. 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 $B$.

  2. 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.

$\endgroup$
5
  • $\begingroup$ So which of OP’s two possible interpretations is correct? $\endgroup$
    – littleO
    Aug 3, 2022 at 17:38
  • $\begingroup$ I think this misses the point of the question. OP says "In the second case, $C$ is not necessarily closed in $X$..." so I think they essentially understand the point you have made. $\endgroup$
    – SBK
    Aug 3, 2022 at 17:42
  • $\begingroup$ Maybe I missed the point of the question. I assumed the question was in the title: "Does the expression 'a closed subset of a set' always imply the subset is closed in the subspace topology?" I think I answered that question. I did not find another question in the body of the text, only background explanation. So, if the title question was not the one to be answered I ask for clarification about the question. $\endgroup$ Aug 3, 2022 at 17:55
  • $\begingroup$ I admit the title was not representing correctly. Thank you. $\endgroup$
    – Hermis14
    Aug 3, 2022 at 18:35
  • $\begingroup$ OK. In that case my answer and I will not be of much help. I am not familiar with the conventions such papers and the associated community use. Apologies. $\endgroup$ Aug 3, 2022 at 22:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .