I am studying the proof of the following theorem, but I don't understand the phrase "we can assume without loss of generality"

Theorem: Let $X$ be a compact space and $Y$ a $KC$ space such that $Y^{\star}=Y\cup \{ \infty \}$ be the one-point compactification of $Y$. Then a set $C\subset X\times Y$ is compactly closed in $X\times Y$ if and only if $C\cup (X\times \{ \infty \})$ is compactly closed in $X\times Y^{\star}$.


  1. $X$ is a $KC$ space if every compact $K\subset X$ is closed.
  2. Let $C\subset X$. Then $C$ is compactly closed if for all closed compact $K\subset X$, if $C\cap K$ is compact.

Proof of the Theorem

Suppose $C$ is compactly closed in $X\times Y$. Let $\Phi=\{V_\alpha \}_{\alpha \in I}$ be an open cover of $C\cup (X\times \{\infty \})$. Since $X\times \{\infty \}$ is compact we can assume without loss of generality that it exists $\alpha_0 \in I$ such that $V_{\alpha_0}$ has the form $V_{\alpha_0}=X\times W$, where $W$ is an open set of $Y^\star$ with $\infty \in W$. By the definition of the topology of $Y^\star$, we have that $W = Y^\star -K$ with $K$ compact and closed subset of $Y$. Note that $Y^\star-W = K$ and therefore $Y^\star-W$ is compact and closed in Y.

Note that $X\times (Y^\star -W)$ is compact and closed in $X\times Y$. Since $C$ is compactly closed in $X\times Y$, we have $H=C\cap (X\times (Y^\star -W) )$ is compact and therefore exists $\{V_{\alpha_1}, V_{\alpha_2}, \ldots , V_{\alpha_n} \} \subset \Phi$ such that $H\subset \bigcup_{i=1}^n V_{\alpha_i}$. Then $C\cup (X\times \{\infty \})\subset \bigcup_{i=1}^n V_{\alpha_i}$

Now suppose $C\cup (X\times \{\infty \})$ is compact in $X\times Y^\star$. Let $K$ be a compact closed subset of $X\times Y$. Consider the projection function $\rho_2: X\times Y \rightarrow Y$. Then we have $\rho_2(K)$ is compact in $Y$. Since $Y$ is a $KC$ space, we have $\rho_2(K)$ is closed in $Y$ and then $\rho_2(K)$ is closed in $Y^\star$. Finally note that $C\cup K=[C\cup(X\times \{\infty \})]\cap (X\times \rho_2(K))\cap K$ is compact and therefore $C$ is compactly closed en $X\times Y$.

If you have any idea why the phrase "we can assume without loss of generality" is valid please let me know

  • 5
    $\begingroup$ I interpret this phrase to mean “we will only prove the special case in which such and such simplifying assumption is true, but the general theorem is an easy corollary of this special case”. $\endgroup$
    – littleO
    Jun 22 at 23:51

1 Answer 1


Given that the family ${\cal V}=\{V_\alpha\}_{\alpha\in I}$ is a cover of $C\cup(X\times\{\infty\})$, it is in particular a cover of $X\times\{\infty\}$, which is compact. Therefore, there exist, say $V_{\alpha_1}, \dots, V_{\alpha_k}$ whose union covers $X\times\{\infty\}$. Thus, if we add their union $U=V_{\alpha_1}\cup\dots\cup V_{\alpha_k}$ to $\cal V$, any finite subcover of the enhanced family ${\cal V}\cup\{U\}$ will ultimately be (equivalent to having) a finite subfamily of $\cal V$. Therefore, there is no loss of generality (in the selection of $\cal V$) in assuming that $U$ was already in $\cal V$, i.e., that there is some $\alpha_0\in I$ such that $X\times\{\infty\}\subseteq V_{\alpha_0}$. Since, moreover, $C\subseteq X\times Y$, we can further assume that $V_{\alpha_0}$ is of the form $X\times W$ for some $W$ open in $Y^*$.


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