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. 

In both cases the definitions are equivalent. It should be clear in each case that if Definition $\mathbf{1}$ is satisfied, then so is Definition $\mathbf{2}$. 
Suppose in the first case that for each open nbhd $U$ of $x$ there is an $m_U\in\Bbb N$ such that $x_n\in U$ whenever $n\ge m_U$, and let $N$ be any nbhd of $x$. By definition there is an open nbhd $U$ of $x$ such that $U\subseteq N$, and clearly $x_n\in N$ whenever $n\ge m_U$.
Finally, suppose in the second case that each open nbhd of $x$ has non-empty intersection with $A$, and let $N$ be any nbhd of $x$. As before, there is an open nbhd $U$ of $x$ such that $U\subseteq N$, and clearly $N\cap A\supseteq U\cap A\ne\varnothing$, so $N\cap A\ne\varnothing$.
