Confusion regarding open and closed subsets with different parent sets.

Question

Consider $A=\{q \in \mathbb Q:q^2 \geq2\}$ and the metric $d(x,y)=|x-y|$.

Then considered as a subset of the metric space $(\mathbb R,d) A$ is not open since none of the points of $A$ is an interior point of A since every nbd in $\mathbb R$ contains irrational numbers not belonging to $A$. Also $\sqrt2 $ is a limit point of $A$ not contained in $A$ so $A$ is not closed either.

However, when considered as a subset of the metric space $(\mathbb Q,d)$ it is both open and closed. Since now the nbds only contain rationals, we can find one of every point that lies inside $A$. And now since the parent space is $\mathbb Q$, $\sqrt 2$ is no longer considered a limit point and the remaining limit points are all the members of $A$ thus lying inside $A$ and making it closed.

Is this reasoning correct? Am I right in assuming that a subset is open or closed depending on the set it is considered a subset of?

Answer 1

Yes, it matters quite a lot which is the parent space when you consider questions of openness or closedness of sets.

In general, if $A \subseteq Y \subseteq X$ then $A$ is closed in $Y$ (where $Y$ has the subspace topology inherited from $X$) iff $A = C \cap Y$, where $C$ is closed in $X$, so closed sets in the subspace are just intersections of closed sets in the larger space with that subspace. The same holds for open sets. In terms of the metric, this nicely corresponds with restricting the metric on $X$ to points of $Y$, as you do here.

$A$ is closed (open) in $\Bbb Q$ because there it is just a closed (open) set of $\Bbb R$ intersected with $\Bbb Q$:, $$A= \left((-\infty,-\sqrt{2}] \cup [\sqrt{2}, +\infty)\right) \cap \mathbb{Q} = \left((-\infty,-\sqrt{2}) \cup (\sqrt{2}, +\infty)\right) \cap \mathbb{Q}$$

nicely abusing the fact that $\pm \sqrt{2}$ are not in $\Bbb Q$.

Answer 2

Yes. A topology $T$ on a set $X$ is a collection of subsets of $X$ such that

$(i)\;\emptyset \in T$ and $X\in T$

$(ii) \; \cup S=\cup_{u\in S}\,u\in T$ whenever $S\subset T$

$ (iii)\; \cap S=\cap_{u\in S}\,u\in T$ whenever $S$ is a $finite$ subset of $T.$

Members of $T$ are called open sets.

This is a very broad definition. The only sets that are guaranteed to belong to $T$ are $\emptyset$ and $X$.

So whether a subset of $X$ is open or closed (or neither ) depends on which $T$ is chosen. A topological space is a pair $(X,T)$ where $T$ is a topology on $X$. It is extremely common to say "the space $X$" although this is technically incorrect.

If $(X,T)$ is a topological space and $Y\subset X$ then $T'=\{t\cap Y:t\in T\}$ is a topology on $Y$. It is extremely common to say "the subspace $Y$" although this is also technically incorrect. Rigorously we should say "the subspace $(Y,T')$ of the space $(X,T)$." And the members of $T'$ are called the open sets of the subspace $(Y,T')$. Whether a subset of the subspace $Y$ (sic) is open or closed ( or neither ) depends on $T$ and $X$.

When considering more than one topology on a set it is best to state these things rigorously to avoid ambiguity.