I have a sequential, hereditarily Lindelöf topological space $\mathcal{X}$, and some subset $A \subseteq \mathcal{X}$. I am interested in the following properties:


1. There is some compact set $B$ with $A \subseteq B \subseteq \mathcal{X}$.
2. Every sequence $(a\_n)\_{n \in \mathbb{N}}$ in $A$ has a subsequence which converges in $\mathcal{X}$.
3. For every open cover $\bigcup\_{n \in \mathbb{N}} U\_n \supseteq \mathcal{X}$ there is some $N \in \mathbb{N}$ with $A \subseteq \bigcup\_{n \leq N} U\_n$.


I know that each of these implies the next, but do the implications reverse?


(Trying to Google sources is made difficult by mentions of relatively compact dominating; and since I don't wish to assume Hausdorffness, that is not the notion I am looking for.)



Being contained in a compact set
It seems that [Gutik's hedgehog](https://mathoverflow.net/questions/296395/what-is-the-genuine-name-for-the-gutik-hedgehog) is a required counterexample.


I recall that Gutik's hedgehog is the set $$X=\{(0,0)\}\cup\{(\tfrac1n,0):n\in\mathbb N\}\cup\{(\tfrac1n,\tfrac1{nm}):n,m\in\mathbb N\}$$
endowed with the topology $\tau$ consisting of subsets $U\subseteq X$ satisfying the following two conditions:


(1) if $(\frac1n,0)\in U$ for some $n\in\mathbb N$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $k\ge m$;


(2) if $(0,0)\in U$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $n\ge m$ and all $k\in\mathbb N$


It is easy to see that Gutik's hedgehog is countable, second-countable, and Hausdorff but not regular.


In Gutik's hedgehog consider the subset $A=\{(0,0)\}\cup\{(\frac 1n,\frac1{nm}):n,m\in\mathbb N\}$.


It is easy to see that every nontrivial sequence in $A$ has a subsequence that converges in $X$ (to a point in the set $\{(0,0)\}\cup\{(\frac1n,0):n\in\mathbb N\}$). On the other hand, no subspace of $X$ containing $A$ is compact. So, the property (2) is satisfied but (1) does not.


The conditions(2) and (3) are equivalent for sequential spaces. This equivalence follows from two lemmas.


**Lemma 1.** Let $A$ be a subset of a topological space $X$ such that every open cover of $X$ contains a finite subfamily covering $A$. Then every sequence in $A$ has an accumulation point in $X$.


*Proof.* Assuming that $A$ contains a sequence $(a\_n)\_{n\in\omega}$ without accumulation points in $X$, for every $x\in X$, we can find a neighborhood $U\_x\subseteq X$ of $X$ such that the set $\{n\in\omega:a\_n\in U\_x\}$ is finite. Then for every finite subfamily $\mathcal F$ of the open cover $\{U\_x:x\in X\}$ of $X$ the set $\{n\in\omega:a\_n\in \bigcup\mathcal F\}$ is finite and hence $\{a\_n:n\in\omega\}\not\subseteq\bigcup\mathcal F$ and $A\not\subseteq \bigcup\mathcal F$. $\quad \square$


**Lemma 2.** If a sequence $(x\_n)\_{n\in\omega}$ in a sequential space $X$ has an accumulation point in $X$, then $(x\_n)\_{n\in\omega}$ has a subsequence that converges to some point of $X$.


*Proof.* Let $a$ be an accumulation point of the sequence $(x\_n)\_{n\in\omega}$ in $X$.


For every point $x\in X$ let $I\_x$ be the intersection of all neighborhoods of $x$ in $X$. It is easy to see that for every $x\in X$ and $y\in I\_x$ we have $I\_y\subseteq I\_x$.


For a subset $A\subseteq X$, consider the set $\dot A:=\{x\in X:I\_x\cap A\ne\emptyset\}$ and observe that $A\subseteq \dot A$.


If for some $x\in X$ the set $\Omega\_x:=\{n\in\omega:x\_n\in I\_x\}$ is infinite, then $(x\_n)\_{n\in\Omega\_x}$ is a required convergent (to $x$) subsequence of the sequence $(x\_n)\_{n\in\omega}$ and we are done. So, assume that for every $x\in X$ the set $\Omega\_x$ is finite. In this case, the set $A=\{x\_n:n\in\omega\setminus\Omega\_a\}$ is not closed in $X$ as $a\in\overline{A}\setminus A$.


Moreover, the set $\dot A$ is not closed as $A\subseteq \dot A\subseteq\bar{A}$ and $a\notin \dot A$. By the sequentiality of $X$, the set $\dot A$ contains a sequence $(a\_k)\_{k\in\omega}$ that converges to some point $b\in X\setminus \dot A$. If $a\_k\in I\_b$ for some $k\in\omega$, then $I\_{a\_k}\subseteq I\_b$ and $a\_k\in \dot A$ imply $I\_b\cap A\supseteq I\_{a\_k}\cap A\ne\emptyset$ and hence $b\in \dot A$, which contradicts the choice of $b$. This contradiction implies that $I\_b\cap\{a\_k:k\in\omega\}=\emptyset$ and hence the set $\{a\_k:k\in\omega\}$ is infinite. Since $\{a\_k:k\in\omega\}\subseteq \dot A$, for every $k\in\omega$ there exists a number $i\_k\in\omega$ such that $x\_{i\_k}\in I\_{a\_k}$. We claim that $b$ is the limit of the sequence $\{x\_{i\_k}:k\in\omega\}$. Indeed, for every open neighborhood $O\_b$ of $b$ in $X$, there exists $m\in\omega$ such that $\{a\_k:k\ge m\}\subseteq O\_b$. Then for every $k\ge m$ we have $x\_{i\_k}\in I\_{a\_k}\subseteq O\_b$, which means that the sequence $(x\_{i\_k})\_{k\in\omega}$ converges to $b$.


Assuming that the set $F=\{x\_{i\_k}:k\in\omega\}$ is finite and taking into account that $b\notin \dot A$, we can find a neighborhood $O\_b$ such that $O\_b\cap F=\emptyset$, which contradicts the convergence of the seqeunce $(x\_{i\_k})\_{k\in\omega}$ to $b$. This contradiction shows that the set $F$ is infinite. Then we can choose an increasing number sequence $(n\_k)\_{k\in\omega}$ such that $F=\{x\_{n\_k}\}\_{k\in\omega}$ and $x\_{n\_i}\ne x\_{n\_j}$ for $i\ne j$. For every $k\in\omega$ find $p\_k\in\omega$ such that $n\_k=i\_{p\_k}$. The convergence of the sequence $(x\_{i\_p})\_{p\in\omega}$ to $b$ implies the convergence of the sequence $(x\_{n\_k})\_{k\in\omega}=(x\_{i\_{p\_k}})\_{k\in\omega}$ to $b$. Therefore, $(x\_{n\_k})\_{k\in\omega}$ is a required convergent subsequence of the sequence $(x\_n)\_{n\in\omega}$. $\quad\square$


