Neighborhood vs open neighborhood I am a beginner in the concept of general Topology. I have some confusion about the utilization of neighborhood and open neighborhood. 
$\mathbf{Definition}$: Let $(X,\mathcal{T})$ be a topological space and $x\in X$. $N\subset X$ is called neighborhood of $x$ if there exists $U\in \mathcal{T}$ such that  $x\in U\subset N$. It is clear that any open set $U\in \mathcal{T}$, such that $x\in U$, is a neighborhood of $x$.
Now, Let $(x_n)_{n\in\mathbb{N}}\subset X$ be a sequence and $x\in X$. 
$\mathbf{Definition\ 1}$: $x$ is a limit of the sequence $(x_n)$ if for any neighborhood $N$ of $x$, there exists $n_0\in\mathbb{N}$ such that $x_n\in N$ for all $n\geq n_0$.
Some people use a different definition:
$\mathbf{Definition\ 2}$: $x$ is a limit of the sequence $(x_n)$ if for any open neighborhood $N$ of $x$, there exists $n_0\in\mathbb{N}$ such that $x_n\in N$ for all $n\geq n_0$.
The only difference is that, in the Definition 1, they consider all neighborhoods of $x$ where in the Definition 2, they only consider open neighborhoods. But is that difference not a problem?
Are those definitions equivalent? The same for the definition of Adherent point.
$\mathbf{Definition\ 1}$: $x$ is an adherent point of $A$ if for any neighborhood $N$ of $x$, $N\cap A\neq \emptyset$.
$\mathbf{Definition\ 2}$: $x$ is an adherent point of $A$ if for any open neighborhood $N$ of $x$, $N\cap A\neq \emptyset$.
I am really confused about those two definitions. 

It should be clear that if $x$ is the limit of $(x_n)$ by Definition $1$, then Definition $2$ is satisfied as well (if it's not clear let me know and I can explain). Hence we only need to show that if Definition $2$ is satisfied, then Definition $1$ is as well:

Suppose $x$ is the limit of $(x_n)$ by Definition $2$. Let $N$ be any neighborhood of $x$, so there exists some open $U$ with $x\in U\subset N$. Because $U$ is an open neighborhood of $x$, by assumption there exists $n_0\in\Bbb{N}$ such that $x_n\in U$ for $n\ge n_0$. But $U\subset N$, so $x_n\in N$ for $n\ge n_0$ as well, and Definition $1$ is indeed satisfied.

So we see the two definitions of a limit are indeed equivalent. I'd recommend you try to apply a similar argument to your second pair of definitions.

Edit: Here is the argument to show Definition $1$ implies Definition $2$:

Suppose Definition $1$ holds. Then if $U$ is any open neighborhood of $x$, because $U$ is a neighborhood Definition $1$ tells us that there is an $n_0$ such that $x_n\in U$ for $n\ge n_0$; since $U$ was arbitrary, we see Definition $2$ holds.

