complete accumulation point Let $(X, ‎\tau‎ )$ be a topological space, which is not compact. Then
there is$ C ⊆ X$, such that $ C$ has no complete accumulation point.
Proof. A space $X$ is not compact, hence there is a strictly increasing infinite
open cover $ \mathcal{U} = \{U_i : i < k \}$. Without loss of generality, we can assume, that $\mathcal{U} $ has the smallest cardinality, i.e.$k$ is an infinite regular cardinal. For any i < k let $x_{i} ∈ U_{i+1} - U_{i}$. The set $C = \{ x_{i} : i < k\}$ has the requested property. Because
if $ x$ is a complete accumulation point of $C$, then every open neighbourhood of $x$
intersects $\{x_{i} : \alpha ≤ i < k \}$ for each $\alpha  < k $, because the complement in $C$ has
cardinality strictly smaller than $ k$. We get, that $x$ can’t be in any $U _{\alpha } $, from
$x ∈ \overline{ \{x_{i} : \alpha  ≤ i < k \} } ⊆ X - U_{\alpha }$ And finally $\mathcal{U}$ doesn’t cover the point x.
(1): Why "every open neighbourhood of $x$
intersects $\{x_{i} : \alpha ≤ i < k \}$ for each $\alpha  < k $, because the complement in $C$ has
cardinality strictly smaller than $ k$.?
(2) Why " $x ∈ \overline{ \{x_{i} : \alpha  ≤ i < k \} } ⊆ X - U_{\alpha }$ And finally $\mathcal{U}$ doesn’t cover the point x?
 A: Since $\kappa$ is a cardinal, then $$C\setminus\{x_i:\alpha\le i<\kappa\}=\{x_i:0\le i<\alpha\}$$ has cardinality at most $|\alpha|,$ which is strictly smaller than $\kappa$. If there were an open neighborhood $U$ of $x$ that didn't intersect $\{x_i:\alpha\le i<\kappa\}$ for some $\alpha<\kappa,$ we would then have $U\cap C\subseteq\{x_i:0\le i<\alpha\},$ but this can't happen, by definition of a complete accumulation point.
Take any $\alpha<\kappa$. Since every open neighborhood of $x$ intersects $\{x_i:\alpha\le i<\kappa\},$ then $x\in\overline{\{x_i:\alpha\le i<\kappa\}}.$ Since the $U_i$ are a strictly increasing chain of sets and since $x_i\notin U_\alpha$ for $i\ge\alpha,$ then $U_\alpha$ and $\{x_i:\alpha\le i<\kappa\}$ are disjoint, and since $U_\alpha$ is open, then $X\setminus U_\alpha$ is a closed superset of $\{x_i:\alpha\le i<\kappa\}.$ Hence, $\overline{\{x_i:\alpha\le i<\kappa\}}\subseteq X\setminus U_\alpha.$
Since for all $\alpha<\kappa,$ $x\notin U_\alpha,$ then $x\notin\bigcup\mathcal U=\bigcup_{\alpha<\kappa}U_\alpha$
