0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.