Lemma 109.4.1. There exists an inverse system of quasi-compact topological spaces over $\mathbf{N}$ whose limit is not quasi-compact.

## 109.4 Non-quasi-compact inverse limit of quasi-compact spaces

Let $\mathbf{N}$ denote the set of natural numbers. For every integer $n$, let $I_ n$ denote the set of all natural numbers $> n$. Define $T_ n$ to be the unique topology on $\mathbf{N}$ with basis $\{ 1\} , \ldots , \{ n\} , I_ n$. Denote by $X_ n$ the topological space $(\mathbf{N}, T_ n)$. For each $m < n$, the identity map,

is continuous. Obviously for $m < n < p$, the composition $f_{p, n} \circ f_{n, m}$ equals $f_{p, m}$. So $((X_ n), (f_{n,m}))$ is a directed inverse system of quasi-compact topological spaces.

Let $T$ be the discrete topology on $\mathbf{N}$, and let $X$ be $(\mathbf{N}, T)$. Then for every integer $n$, the identity map,

is continuous. We claim that this is the inverse limit of the directed system above. Let $(Y, S)$ be any topological space. For every integer $n$, let

be a continuous map. Assume that for every $m < n$ we have $f_{n,m} \circ g_ n = g_ m$, i.e., the system $(g_ n)$ is compatible with the directed system above. In particular, all of the set maps $g_ n$ are equal to a common set map

Moreover, for every integer $n$, since $\{ n\} $ is open in $X_ n$, also $g^{-1}(\{ n\} ) = g_ n^{-1}(\{ n\} )$ is open in $Y$. Therefore the set map $g$ is continuous for the topology $S$ on $Y$ and the topology $T$ on $\mathbf{N}$. Thus $(X, (f_ n))$ is the inverse limit of the directed system above.

However, clearly $X$ is not quasi-compact, since the infinite open covering by singleton sets has no inverse limit.

**Proof.**
See discussion above.
$\square$

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