Compactness and sequential compactness are different in general topological spaces. However, in a metric space, they are equivalent. The weakest property that is equivalent to compactness in metric spaces that I can think of is pseudocompactness (i.e. every continuous real-valued function is bounded). Because all the other properties that are equivalent to compactness such as sequential compactness or countable compactness imply pseudocompactness. So I'm wondering if there are other properties equivalent to compactness in metric spaces that do not imply pseudocompactness?
Which topological properties are equivalent to compactness in metric spaces
compactnessgeneral-topologymetric-spaces
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I already showed in this answer that $X'$ relatively compact implies $X'$ relatively countably compact. Also, if $X'$ is relatively compact, the $\overline{X'}$ is compact, and thus sequentially compact (this holds in particular in metric spaces, but also more broadly). So any sequence from $X'$ has a convergent subsequence with limit in $\overline{X'}$, so $X'$ is then relatively sequentially compact as well.
In any space, $X'$ relatively sequentially compact implies $X'$ relatively countably compact: any infinite subset $A$ of $X'$ contains some sequence with all different elements, which has a convergent subsequence to some $x \in X$, and this $x$ is an accumulation point of $A$.
If $X'$ is relatively countably compact, and $X$ is metric (first countable and $T_1$ will already do), let $(x_n)$ be a sequence from $X'$. If $A = \{x_n: n \in \mathbb{N}\}$ is finite, some value occurs an infinite number of times, and yields a convergent subsequence. So assume $A$ is infinite, so it has an accumulation point $p \in X$. Because $X$ is $T_1$, this means that every neighbourhood of $p$ intersects $A$ in infinitely many points. Pick $x_{n_1}$ in $B(p, 1)$, $x_{n_2}$ with $n_2 > n_1$ in $B(p, \frac{1}{2})$, and so on, by recursion. This defines a convergent subsequence of $(x_n)$ that converges to $p$. So $X'$ is relatively countably compact.
If $X'$ is relatively countably compact, and $X$ is metric, then we do get that $X'$ is relatively compact (this is due to Hausdorff, IIRC). I cannot reconstruct a proof right away, but there is one here, e.g. (but this involves Cauchy filters etc.)
It turns out that these notions are also equivalent in some topological vector spaces (spaces of the form $C_p(X)$ where $X$ is compact (Grothendieck), and weak topologicals on normed vector spaces (Eberlein-Smulian)), but these are a bit more involved, I think.
No, they're not. The standard examples are all completely regular, so uniformisable. e.g.
$\omega_1$ (countably compact, sequentially compact, not compact).
$\{0,1\}^\mathbb{R}$ (compact, countably compact, not sequentially compact)
Of course, compact always implies countably compact in all spaces, we have a "free" implication there. To get the equivalence of sequential and countable compactness (both essentially "countable" notions) we need something like sequentiality of $X$ (so first countability of metric spaces is a reason), so that sequentially closed and closed sets are the same. $T_1$-ness is also handy there (but this is included in Tychonoffness/uniformisability, at least in my definition).
I don't think that uniform structure has much to do with things like sequential compactness (being a sequential space is independent of that). To go from countable compactness to compactness, Lindelöfness is necessary and sufficient, trivially. This also has nothing to with uniform structure. There we'd have to go to properties like "uniform boundedness" and "completeness", two uniform space notions that together are equivalent to compactness (where completeness is meant in the nets or filter sense, not sequential completeness as in metric spaces). You might want to look into that direction for equivalences.
Best Answer
The general topology properties
are all equivalent in metrisable spaces, but potentially different in general spaces.
And a uniform spaces characterisation: complete plus totally bounded (these notions do not apply to all topological spaces so I mention this separately), or the fact that every compatible metric is bounded or that every compatible metric is complete (notions only applicable to metrisable spaces).