What is the neighborhood-definition of a compact space? Looking through some articles on topology, I see a neighborhood definition of 'topological space' and a neighborhood definition of 'Hausdorff space', but do not see a neighborhood definition of 'compact space'.
The 'open cover' definition of compact space states it is where each open cover has a finite subcover. Is this equivalent to a space being compact when every union of neighborhoods has a finite subcover consisting of neighborhoods?
What is the neighborhood definition of a compact space?
 A: 
The 'open cover' definition of compact space states it is where each open cover has a finite subcover. Is this equivalent to a space being compact when every union of neighborhoods has a finite subcover consisting of neighborhoods?

This depends on what you mean exactly. There are three possible specific definitions that comes to mind:

Let $(X, \mathcal N)$ be a neighbourhood space.

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*$X$ is α-compact, if for every covering $X = \bigcup_{i ∈ I} N_i$ of neighbourhoods, such that for all $i ∈ I$ there is some $x ∈ X$ with $N_i ∈ \mathcal N(x)$, there is a finite subcovering; that is, there is finite some subset $J ⊆ I$ such that $X = \bigcup_{i ∈ J} N_i$.

*$X$ is β-compact, if for every covering $X = \bigcup_{c ∈ C} N_c$ of neighbourhoods, such that for all $x ∈ X$ there is some $c ∈ C$ with $N_c ∈ \mathcal N(x)$, there is a finite subcovering; that is, there is finite some subset $D ⊆ C$ such that $X = \bigcup_{c ∈ D} N_c$.

*$X$ is γ-compact, if for every covering $X = \bigcup_{x ∈ X} N_x$ of neighbourhoods, such that for all $x ∈ X$ we have $N_x ∈ \mathcal N(x)$, there is some finite subcovering of $X$; that is, there is some finite subset $T ⊆ X$ such that $X = \bigcup_{x ∈ T} N_x$.


Let’s check these definitions. Obviously, we have $α \implies β \implies γ$.

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*α. The compact space $[0..1]$ is covered by the sets $\{0\} ∪ [1/(n+1)..1/n];~n ∈ ℕ$, all of which are neighbourhoods of some point. However you can’t leave out any of the set, lest you are missing some segment $[1/(n+1)..1/n]$. So $[0..1]$ isn’t α-compact.

Now, it turns out, that both $β$ and $γ$ are equivalent to being compact. We show
$$γ \implies \text{compactness} \implies β.$$

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*“$γ \implies \text{compactness}$”. Let $X = \bigcup_{i ∈ I} U_i$ be an open cover of $X$. For each $x ∈ X$ choose some $i_x ∈ I$ with $x ∈ U_{i_x}$. Now obviously $X = \bigcup_{x ∈ X} U_{i_x}$. By γ, there is some finite $T ⊆ X$ with $X = \bigcup_{x ∈ T} U_{i_x}$, giving a finite subcover of $X = \bigcup_{i ∈ I} U_i$.

*“$\text{compactness} \implies β$”. Let $X = \bigcup_{c ∈ C} N_c$ such that for all $x ∈ X$ there is some $c_x ∈ C$ with $N_{c_x} ∈ \mathcal N(x)$. Then for all $x ∈ X$, we have $x ∈ N_{c_x}^\circ$, so $X = \bigcup_{x ∈ X} N_{c_x}^\circ$. As $X$ is compact, there is a finite subcover of this, yielding a finite subcover $X = \bigcup_{c ∈ D} N_c$ for some finite $D ⊆ C$.

Conclusion. You can use both β and γ as a neighbourhood-definition for compactness, the definition $γ$ being more beautiful and the definition β being more widely applicable.
A: Let $T$ be a topological space.


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*Neighborhood definition of open set:


A set $S \subset T$ is open means every $s \in S$ has a neighborhood $N_s$ contained in $S$.
That is, $\forall s \in S$ $\exists N_s \subset S$.


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*Open set definition of open cover:


Given a set $S$ in a topological space, an open cover of $S$ is a family of open sets $O_i |_{i \in I}$ whose union contains $S$. That is; $O_i |_{i\in I}$ is an open cover of $S$ means $O_i$ is open $\forall i \in I$ and $S \subset \bigcup_{i \in I} O_i$.


*

*Open cover definition of compact space:


$T$ is compact means every open cover has a finite subcover. That is, given an open cover $O_i |_{i\in I} \subset T$ $\exists$ finite $J \subset I$ such that $O_j |_{j\in J} \subset T$.

Thus, neighborhoods form a definition of open set, which forms the definition of open cover, which forms the definition of compact space.
To describe a compact space directly in terms of neighborhoods, we might say:
A topological space $T$ is compact means given a covering family $T \subset O|_{i \in I}$ where $O_i$ contains a neighborhood for each of its points, there is a finite subfamily that also covers $T$. That is, Given $O|_{i \in I}$ where $T \subset O|_{i \in I}$ and $\forall x \in O_i$ $\exists N_x \in O_i$, $\exists$ finite $J \subset I$ such that $T \subset O|_{j \in J}$.
