Find a point $l$ in the closure of $A$ so that no sequence with values in $A$ converges to $l$. Find an example for this. You can choose which space to use. An example the professor gave had to do with the Box Topology, but I was wondering if there was an easier example. Thanks.
 A: One example is $X=(\Bbb R,\tau)$, where $\tau$ is the cocountable topology, that means the closed sets are exactly the countable sets and $\Bbb R$ itself. Note that the convergent sequences in $X$ are eventually constant. Here we can find a limit point of a set which is not the limit of a sequence within that set. In fact, we have an even more extreme situation here: The set $A=(-\infty,0]$ is not closed, as its closure is all of $\Bbb R$, but each sequence within $A$ has its limit in $A$. (Spaces without such sets are called sequential spaces). So each limit point of $A$ outside of $A$ is not limit of any sequence in $A$. In particular, there is a limit point which is not a sequential limit, and spaces where this can not happen are called Fréchet-Urysohn spaces.
There are also sequential spaces which are still not Fréchet-Urysohn, but they are all characterized by containing one very special space which is called the Arens space
A: Here is a countable example. Define a topology on the set $\mathbb N$ of all natural numbers by calling a set $U$ open if either $1\not\in U$ or else $\sum_{n\not\in U}\frac1n\lt\infty$. Then $1$ is in the closure of the set $A=\mathbb N\setminus\{1\}$, but no sequence with values in $A$ converges to $1$.
A: It seems like you need to have a space where there's a point that's infinitely close to $A$, but far from any countable collection.  So, let $A$ be $\{ \alpha\, |\, \alpha < \omega_1 \}$, all ordinals less than or equal to the first uncountable ordinal, and take the topology generated by all sets $\{ \alpha\, |\, \alpha < \lambda \}$ for some ordinal $\lambda$.  Then $\omega_1$ is in the closure of $A$, but for any countable ordinal $\lambda$, $\{\alpha\, |\, \alpha\, \leq \lambda \}$ is closed, so its complement is an open neighborhood of $\omega_1$.  Since any sequence in $A$ consists of a countable collection of countable ordinals, it's bounded by a countable ordinal $\lambda$, so the complement of $\{\alpha\, |\, \alpha \leq \lambda \}$ is an open neighborhood of $\omega_1$ disjoint from the sequence.  
A: The box topology is one of the most elementary ways to produce an example, but there are many others; which are easy depend on what you know.
Here’s one that I think is both elementary and easy. Define a new topology $\tau$ on $\Bbb R$ by making each point of $\Bbb R\setminus\{0\}$ isolated and saying that $U\subseteq\Bbb R$ is a nbhd of $0$ if and only if $0\in U$ and $\Bbb R\setminus U$ is countable. $\Bbb R\setminus\{0\}$ is uncountable, so every nbhd of $0$ intersects $\Bbb R\setminus\{0\}$, and therefore $0\in\operatorname{cl}_\tau(\Bbb R\setminus\{0\})$. However, no sequence in $\Bbb R\setminus\{0\}$ converges to $0$: if $\langle x_n:n\in\Bbb N\rangle$ is any sequence in $\Bbb R\setminus\{0\}$, $\Bbb R\setminus\{x_n:n\in\Bbb N\}$ is a nbhd of $0$ that contains no term of the sequence at all.
Every ordinal space $\alpha$ with $\alpha>\omega_1$ contains an example, as does every non-trivial Čech-Stone compactification $\beta X$, and in particular the space $\beta\omega$ (or if you prefer, $\beta N$).
A: I have found a relatively technical example, but I think could be fun to share it with you.
Let's first state a
$\mathbf{Theorem:}(Schur)$
In the Banach space $\ell_1$ the $\Vert{\cdot}\Vert_{\ell_1}$ convergence and the weak convergence are equivalent.
Define now $B:=\{ x\in\ell_1,\text{such that } \Vert x \Vert\leq 1 \}$ and $S:=\{ x\in\ell_1,\text{such that } \Vert x \Vert = 1 \}$.
Observe that in such spaces $S$ is weakly dense in $B$ (this can be easily seen as each open set that intersect $B$ contains a subspace of finite co-dimension, namely the intersection of kernels of a finite number of functionals), thus $S$ is not weakly closed in $B$.
Take now a weekly converging sequence in $S$ then the limit must be in $S$
(as $\Vert{\cdot}\Vert_{\ell_1}$ is a continuous function and for $Schur's\  theorem$ the sequence is also norm converging).
This shows that $S$ is sequentially closed in the weak topology but not weakly closed.
Now just take $y\in B\setminus S$ and you have your point.
