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 $E' \not\subset E$


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?


Your proof is correct except the conclusion. What you have proved is the $E$ must be simultaneously open and closed.

  • $\begingroup$ @ArinCharudhuri So instead of having a proof by contradiction, I actually had a direct proof, that $E$ must be simultaneously open and closed? $\endgroup$ – Perturbative Oct 16 '16 at 0:52
  • $\begingroup$ Yes. You are correct. $\endgroup$ – Arin Chaudhuri Oct 16 '16 at 0:54

To say that a set $S\subseteq X$ is open in $(X,d)$ does not mean $S$ is not closed in $(X,d)$. Similarly, To say that $S$ is closed in $(X,d)$ does not mean $S$ is not open in $(X,d)$. "Closed" is not defined to be "not open", neither is "open" defined to be "not closed".

You can find a simple example showing that an open set can contain all its limit points. Take the whole space $X$. The set of all limit points, by definition, does not go outside of $X$, so that $X'\subseteq X$. Also, $X$ is open in $(X,d)$, so that it is open and contain all its limit points.


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