Problem understanding a proof in Model Categories by Hovey I have serious problems understanding this proof from the book Model Categories, by Mark Hovey:


Here's a list of things I don't understand:


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*He's trying to prove the assertion by contradiction, by showing that there is a sequence in $f(A)$ (which is compact) with no limit point. Thus he should prove that $S$ is discrete in the subspace topology and is closed as a subset of $f(A)$ (only discreteness wouldn't be sufficient, indeed the sequence $\{1/n\}_{n\in\mathbb{N}}\subset\mathbb{R}$ has the discrete topology as a subspace of $\mathbb{R}$ and $0$ as a limit point). Hovey claims to have proven discreteness, but I don't seeany proof of closeness.

*I don't understand why $S$ should have the discrete topology as a subspace of $X_\mu$.


Any help in understanding the proof would be gratly appreciated.

NOTES:


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*A closed $T_1$ inclusion is defined as an inclusion $f:X\rightarrow Y$ (so, $U\subseteq X$ is open if, and only if, there is some $V\subseteq Y$ such that $f^{-1}(V) = U$) which is also a closed map as such that every point in $Y\backslash f(X)$ is closed.

 A: The assumption that the $X_\alpha$ form a $\lambda$-sequence includes in particular that for every limit ordinal $\kappa \leq \lambda$ we have $X_\kappa = \varinjlim_{\alpha \lt \kappa} X_\alpha$.
Since directed colimits are unchanged by passing to cofinal subsets of the indexing set, we have $X_\mu = \varinjlim_{\alpha_n} X_{\alpha_n}$. In other words, $X_\mu$ carries the final topology  induced by the system $X_{\alpha_0} \to X_{\alpha_1} \to \cdots$. This implies that a subset $B$ of $X_\mu$ is closed iff $X_{\alpha_n} \cap B$ is closed in $X_{\alpha_n}$ for all $n$. 
Hovey argues that for every subset $S' \subseteq S$ we have that $X_{\alpha_n} \cap S'$ is closed in every $X_{\alpha_n}$ (since it is finite and contained in $X_{\alpha_n} \setminus X_{\alpha_0}$), so $S'$ is closed in $X_\mu$. This applies in particular to $S$ itself, so $S$ is closed in $X_{\mu}$ and since every subset $S' \subseteq S$ is closed in $X_\mu$, it follows that $S$ is discrete.
On the other hand, the map $X_\mu \to X_\lambda = \varinjlim_{\alpha \lt \lambda} X_\alpha$ is a closed inclusion, so every subset $S'$ of $S$ is also closed in $X_\lambda$. Since every subset of $S$ is closed in $X_\lambda$, we have that $S$ is an infinite closed discrete subset of the compact space $f(A) \subseteq X_\lambda$, which is impossible.
A: 1) I will prove a general result:
Result 1: Let $X$ be a $T_1$ space and $S$ be a subset of $X$. If $S$ hasn't an accumulation, then it is closed and discrete.
Proof: $S$ is closed: since $\overline{S}=S\cup S'$ and $S'=\emptyset$, then $\overline{S}=S$. $S$ is discrete: for any point $s\in S$, there exists an open set $U$ of $s$ such that $U\cap S$ is finite (Why?), say $\{s_i: 1\le n\}$, and hence $s$ has an open set $O=U \cap \{U_i:i\le n\}$ such that $O\cap S=\{s\}$, which witnesses that $S$ is discrete, where each $U_i$ is an nbhd of $s$ such that $x_i \notin U_i$. This complete the proof.
2) Since $S$ is discrete, there exaist $\{U_s: s\in S\}$ of $f(A)$ such that $U_s \cap S=\{s\}$. To prove that $S$ is discrete in the $X_\mu$ as the subsapce of $X_\mu$. Just let $V_s=U_s \cap X_\mu$. Then $\{V_s: s\in S\}$ is the family of open sets of $X_\mu$ witnesses that $S$ is discrete in the $X_\mu$.
