Let's say I have a set $E$ as a subset of a metric space $X$
If $E$ is open, then is the set of all limit points of $E$, (which we'll denote $E'$) a subset of $E$?
I attempted to prove that if $E$ is open, then $E' \not\subset E$
Proof:
Let $E$ be open. Assume $E' \subset E$.
The closure of $E$, is $\overline{E} = E \cup E'$.
But since $E' \subset E$, we have $a \in E' \implies a \in E$
$$ \therefore E \cup E' = E$$
and thus we have $$\overline{E} = E $$
which contradicts the fact that $E$ is open.
Therefore we can conclude that for any open set $E$, the set of all limit points $E'$ is not contained in $E$, i.e. $E' \not \subset E$. $\ \ \square$
Firstly is my above proof incorrect? If not then the thing is that there can be metric spaces which are both open and closed, take $\mathbb{R^2}$ for example. And if we let $E = \mathbb{R^2}$, then the above proof says that $\mathbb{R^2}$ is closed and not open.
I've heard something about ambient spaces, which is supposed to be the space containing all spaces you are considering, in this case $X$ would be an ambient space, and $E$ would not be an ambient space.
Does the concept of ambient spaces affect whether a set can be open, closed or both? For example if we let $X = \mathbb{R^3}$, and $E = \mathbb{R^2}$, where $E \subset X$, then is $E$ open, closed or both open and closed?
If it does, then does that mean that a metric space can only be open/closed or both, relative to itself or some other metric space which acts as an ambient space?
Best Answer
Your proof is correct except the conclusion. What you have proved is the $E$ must be simultaneously open and closed.