Equivalent properties for a kind of relative compactness? Let $X$ be a topological space and $Y \subseteq X$.
Consider the following statements:
(i) Every net in $Y$ has a cluster point in $X$.
(ii) Every infinite subset of $Y$ has a complete accumulation point in $X$.
These are two possible definitions for relative compactness of $Y$ in $X$.
Question 1: Is there some relation between (i) and (ii)? Does it hold (i) $\Leftrightarrow$ (ii)?
I'm unfamiliar with the formal set theoretical considerations, but I would try to "prove" as follows.
"(i) $\Rightarrow$ (ii)": Let $Z \subseteq Y$ infinite and let $\alpha$ denote the least ordinal such that $|\alpha| = |Z|$ (for which it then follows that $\alpha = |\alpha|$ when cardinals are represented by ordinals), i.e. we have a bijection $x : \alpha \to Z$ which gives us a net $x_a$ such that $\{ x_a \mid a \in A \} = Z$. Now $x_a$ has a cluster point $x \in X$ and show that $x$ is a complete accumulation point of $Z$. Herefore, let $U$ be an open neighborhood of $x$. We know that $x_a$ is frequently in $U$ which means that for each $a_0 \in \alpha$ there is $a \geq a_0$ such that $x_a \in U$. But from this fact, I do not see, why it should hold that $|U \cap \{ x_a \mid a \in A \}| = |U|$. (Why cannot it be the case that if $x_a \in U$ for some $a$ and all $x_b \not\in U$ for $b < a$ that we get a gap in the cardinality? Does it has to do something with the fact that a strictly decreasing sequence of ordinals stabilizes after finitely many steps?)
"(ii) $\Rightarrow$" (i): Let $(x_a)_{a \in A}$ be a net in $Y$ and set $Z := \{ x_a \mid a \in A \}$. If $Z = \{ y_1, \dots, y_n \}$ is finite then some point $y_k \in Z$ is a cluster point. (If not, then for each $k$ there exists an open neighborhood $U_k$ of $y_k$ and $a_k \in A$ such that for all $a \geq a_k$ it holds $x_k \not \in U_k$. Since $A$ is directed the finite set $\{ a_1, \dots, a_k \}$ has an upper bound $s$ and it follows that $x_s \not \in \bigcup_k U_k$ which is a contradiction.) So assume that $Z$ is infinite. Then $Z$ has a complete accumulation point $x \in X$. If $x$ is not a cluster point of $x_a$ then there is an open neighborhood $U$ of $x$ and $a_0 \in A$ such that for all $a \geq a_0$ it holds $x_a \not\in U$. However, we have that $|U \cap Z| = |Z|$ which implies $|U \cap \{ x_a \mid a < a_0 \}| = | \{ x_a \mid a \in A \}|$, so I have a bijection of $\{ x_a \mid a \in A \}$ onto its subset $U \cap \{ x_a \mid a < a_0 \}$. So our situation is as follows: We have a partition $Z = Z_1 \cup Z_2$ with $Z_1 := \{ x_a \mid a \geq a_0 \} \subseteq X \setminus U$ and $Z_2 := \{ x_a \mid a < a_0 \} \cap U$ such that $Z$ and $Z_2$ have the same cardinality, and thus $|Z_1| \leq |Z_2|$. I do not see any possible contradiction.
Question 2: So if there is no relation between (i) and (ii) are there some relations to other relative compactness properties (e.g. $\overline{Y}$ is compact or $Y$ is contained in a compact subset)?
 A: Here’s a partial answer.
Let $Z$ be an infinite subset of $Y$, let $\kappa=|Z|$, and suppose that $Z$ has no complete accumulation point in $X$. Then each $x\in X$ has an open nbhd $U_x$ such that $|U_x\cap Z|<\kappa$. Let 
$$\mathscr{F}=\{F\subseteq Z:|Z\setminus F|<\kappa\}\;;$$
$\mathscr{F}$ is a filter on $Z$. Let
$$D=\{\langle F,z\rangle\in\mathscr{F}\times Z:z\in F\}\;,$$
and for $\langle F_0,z_0\rangle,\langle F_1,z_1\rangle\in D$ write $\langle F_0,z_0\rangle\preceq \langle F_1,z_1\rangle$ if and only if $F_0\supseteq F_1$; then $\langle D,\preceq\rangle$ is a directed set, and
$$\nu:D\to Y:\langle F,z\rangle\mapsto z$$
is a net in $Y$. 
Let $x\in X$ be arbitrary. Let $F_0=Z\setminus U_x$, and let $z_0\in F$ be arbitrary; $|Z\setminus F_0|=|U_x\cap Z|<\kappa$, so $\langle F_0,z_0\rangle\in D$. But if $\langle F,z\rangle\in D$ and $\langle F_0,z_0\rangle\preceq\langle F,z\rangle$, then $z\in F\subseteq F_0=Z\setminus U_x$, so $\nu$ is not frequently in $U_x$. Thus, $x$ is not a cluster point of $\nu$, and $\nu$ therefore has no cluster point in $X$. This shows that (i) implies (ii).
I’ll have to give the opposite implication more thought, though I’m inclined to suspect that it’s false in general.
