Describing the topology of "Alexandroff" compactification Let $(X,\tau)$ and $(A(X)=X\cup \{\infty \},\tau ')$ be two topological spaces. 
I've seen two ways of writing the topology of the Aleksandrov compactification.
$$\tau' =\tau \cup \{A(X)\setminus C\mid C \text { is compact in }X\}$$
$$\tau' =\tau \cup \{(X\setminus C)\cup \{\infty\} \mid C \text { is compact and closed in }X\}$$
Thoughts: Firstly I notice that the following set identity holds $$(A\cup B) \setminus C=(A\setminus B ) \cup (B\setminus C)$$
Hence, $(X\cup \{\infty\})\setminus C=(X \setminus C) \cup (\{\infty\}\setminus C)$ and since $\{\infty\}$ and $C$ are disjoint it follows that $(X\cup \{\infty\})\setminus C=(X \setminus C) \cup (\{\infty\})$. 
Question: What I didn't understand is how to compare the requirements for $C$. 
Note: Alexandroff isn't transliterated correctly. One should write Aleksandrov (from Александров) so please do not interpret this as a typo. Sadly most of the russian names in the mathematical literature are transliterated incorrectly. 
Added: These definitions are indeed the same if we require $X$ to be Hausdorff. It follows from the fact that a compact subset $A \subset X$ of a Hausdorff space $X$ is closed. Usually, when one studies this topic one reaches the conclusion that if we have a topological space $X$ that is locally compact and Hausdorff (but not compact) then the Aleksandrov compactification is unique up to homeomorphism and turns $X$ into a compact and Hausdorff space while preserving the fact that $X$ is homeomorphic to the subspace $A(X)\setminus \{\infty\}$. In this context clearly both definitions work.  
 A: The correct general definition is $C$ should be closed and compact, because the idea of a compactification like $\alpha X$ of $X$ is that $X$ as a subspace of the compactification $\alpha X = X \cup \{\infty\}$ has the same topology as $X$ originally had. This means that $(X \setminus C) \cup \{\infty\}$, which is the “new”  kind of open set we introduce for $\alpha X$ should obey the property that $((X \setminus C) \cup \{\infty\}) \cap X =X \setminus C$ is already open in $X$, or else the subspace topology on $X$ from $\alpha X$ would be strictly finer than the topology on $X$ we started with. This thus implies that $C$ should be closed (as well as compact, which is needed to make $\alpha X$ compact).
We lose nothing by this, because in the most common case where $X$ is Hausdorff the condition is redundant, and the subspace property holds anyway, but for general spaces where compact sets need not be closed, the distinction matters, because otherwise we don’t have a compactification in the sense that we do not change the topology on $X$.
