The closure is a subset of a closed set - Metric spaces

Question

I'm trying to show that if $X$ is a metric space and $E, F \subset X$ where $F$ is a closed set and $E \subset F$ then we have $\bar{E} \subset F.$

Where $\bar{E}$ dentotes the closure of the set $E$. In other words $\bar{E} = E \cup E'$.

$E'$ denotes the set of all limit points.

Attempt:

If $F$ is closed and $ E \subset F$, then $F' \subset F.$ This follows since $F$ is closed. If I can show that $E' \subset F$ then I am done. I'm note quite sure how to do this. I can't just say that since $ E \subset F$ it is closed, as that is false.

Any hints appreciated.

Thanks

Answer 1

$A \subset B$ implies $A' \subset B'$ by definition of limit points. Hence $E' \subset F'$ and $F' \subset F$ so $E' \subset F$.

Answer 2

HINT

Use the sequential characterisation of a closet : a subset $A$ of a metric space $X$ is closed if and only if all convergent sequences of elements of $A$ has its limit in $A$.

SOLUTION (SPOILERS)

Let $(x_n)$ be a sequence of elements of $E$ which converges to $x \in X$. Since $E \subset F$, $(x_n)$ is a sequence of elements of $F$ and since $F$ is closed we have $x \in F$.

Therefore $\bar{E} \subset F$.

Answer 3

This result holds in any topological space, not just metric spaces.

Let $X$ be any topological space, let $F$ be any closed subset of $X$, and let $E$ be any set such that $E \subset F$. Then $\overline{E} \subset F$ also.

Proof:

Let $p$ be any limit point of set $E$ in $X$. Let $U$ be any open set containing $p$. Then $$U \cap E \setminus \{ p \} \neq \emptyset. $$ That is, there exists a point $x \in X$ such that $x \neq p$ and $x \in U \cap E$. But as $E \subset F$, so this $x$ also satisfies $x \in U \cap F$. Thus we obtain $x \in U \cap F \setminus \{ p \}$, which shows that $$ U \cap F \setminus \{ p \} \neq \emptyset, $$ where $U$ is any open set of $X$ containing point $p$. So $p$ is also a limit point of set $F$.

Thus we have shown that $$ E^\prime \subset F^\prime. \tag{1} $$

But as set $F$ is closed in $X$, so we also have $$ F^\prime \subset F. \tag{2} $$

From (1) and (2) we also obtain $$ E^\prime \subset F, $$ and as $E \subset F$, so we have $$ \overline{E} = E \cup E^\prime \subset F \cup F = F, $$ that is, $$ \overline{E} \subset F, $$ as required.

Hope this helps.

Answer 4

Another definition of closure of a set is the following.

Let be $E$ a set then $\bar{E}=\cap_{\gamma \in \Gamma} D_{\gamma}$ where $\{ D_{\gamma} :$ are closed sets that contain $E$)

Then as $F$ is closed and contain $E$, you have $\bar{E} \subset \cap_{\gamma \in \Gamma} D_{\gamma} \subset F$.