Recurrent point: two definitions Let $X$ be a topological space and $T:X\longrightarrow X$ a function. Now lets look at the two following definitions:

  
*
  
*$x\in X$ is a recurrent point if for every neighborhood $U$ (of $x$) the set $$\{n\in\mathbb N\,:\, T^n(x)\in U\}$$ has infinite
  cardinality.
  
*$x\in X$ is a recurrent point if for every neighborhood $U$ (of $x$) there is some $n\in\mathbb N$ such that $T^n(x)\in U$
  

I think that these two definitions are equivalent only when $X$ is a first countable space, because in order to show  $2\Rightarrow 1$, I need to construct a decreasing sequence of neighborhoods that converges to $x$. Is this reasoning right, or the equivalence is true also for a generic $X$?
Edit: Condition $1.$ was edited because there was a typo.
Thanks in advance.
 A: Generally, condition 2 does not imply condition 1.
The criterion is however not first countability.
Suppose $X = \{x,y,z\}$, with the topology $\{ \varnothing, \{x,y\},X\}$, and $T(x) = y;\, T(y) = T(z) = z$. Then $x$ is a recurrent point according to definition 2, but not according to definition 1.
However, if $X$ is a $T_1$-space, the two definitions are equivalent.
For, if $x$ is not a recurrent point according to definiton 1, there is a neighbourhood $U$ of $x$ such that $A = \{n\in\mathbb{N} : T^n(x)\in U\}$ is finite, then we have $T^n(x)\neq x$ for all $n\in \mathbb{N}$ (otherwise $A$ could not be finite), and
$$V = U\setminus \left\{ T^n(x) : n\in A\right\}$$
is a neighbourhood of $x$ with $T^n(x) \notin V$ for all $n\in\mathbb{N}$, so $x$ is also not a recurrent point according to definition 2.
A: With the definition as it's currently written, $2 \implies 1$ as long as single points are closed (I think this is called the $T_0$ property?).
In order to show that this is the case:
If $T^n(x) = x$ for some $n \in \Bbb N$, then $\{T^n(x)\}_{n \in \Bbb N}$ is a periodic orbit so that both 1 and 2 are satisfied.
Suppose then that $x$ satisfies condition 2, but that the above is not the case.  Take an arbitrary neighborhood $U = U_1$ of $x$.  Let $x_1 = T^n(x) \in U$, as guaranteed by 2.  Let $U_2 = U \setminus\{x_1\}$.  Select a suitable $x_2 \in U_2 \subset U$, and set $U_3 = U_2 \setminus \{x_2\}$.  Repeat ad infinitum, and deduce that 1 must hold.
