Definition 5.13.1. A topological space $X$ is called *locally quasi-compact*^{1} if every point has a fundamental system of quasi-compact neighbourhoods.

## 5.13 Locally quasi-compact spaces

Recall that a neighbourhood of a point need not be open.

The term *locally compact space* in the literature often refers to a space as in the following lemma.

Lemma 5.13.2. A Hausdorff space is locally quasi-compact if and only if every point has a quasi-compact neighbourhood.

**Proof.**
Let $X$ be a Hausdorff space. Let $x \in X$ and let $x \in E \subset X$ be a quasi-compact neighbourhood. Then $E$ is closed by Lemma 5.12.4. Suppose that $x \in U \subset X$ is an open neighbourhood of $x$. Then $Z = E \setminus U$ is a closed subset of $E$ not containing $x$. Hence we can find a pair of disjoint open subsets $W, V \subset E$ of $E$ such that $x \in V$ and $Z \subset W$, see Lemma 5.12.4. It follows that $\overline{V} \subset E$ is a closed neighbourhood of $x$ contained in $E \cap U$. Also $\overline{V}$ is quasi-compact as a closed subset of $E$ (Lemma 5.12.3). In this way we obtain a fundamental system of quasi-compact neighbourhoods of $x$.
$\square$

Lemma 5.13.3 (Baire category theorem). Let $X$ be a locally quasi-compact Hausdorff space. Let $U_ n \subset X$, $n \geq 1$ be dense open subsets. Then $\bigcap _{n \geq 1} U_ n$ is dense in $X$.

**Proof.**
After replacing $U_ n$ by $\bigcap _{i = 1, \ldots , n} U_ i$ we may assume that $U_1 \supset U_2 \supset \ldots $. Let $x \in X$. We will show that $x$ is in the closure of $\bigcap _{n \geq 1} U_ n$. Thus let $E$ be a neighbourhood of $x$. To show that $E \cap \bigcap _{n \geq 1} U_ n$ is nonempty we may replace $E$ by a smaller neighbourhood. After replacing $E$ by a smaller neighbourhood, we may assume that $E$ is quasi-compact.

Set $x_0 = x$ and $E_0 = E$. Below, we will inductively choose a point $x_ i \in E_{i - 1} \cap U_ i$ and a quasi-compact neighbourhood $E_ i$ of $x_ i$ with $E_ i \subset E_{i - 1} \cap U_ i$. Because $X$ is Hausdorff, the subsets $E_ i \subset X$ are closed (Lemma 5.12.4). Since the $E_ i$ are also nonempty we conclude that $\bigcap _{i \geq 1} E_ i$ is nonempty (Lemma 5.12.6). Since $\bigcap _{i \geq 1} E_ i \subset E \cap \bigcap _{n \geq 1} U_ n$ this proves the lemma.

The base case $i = 0$ we have done above. Induction step. Since $E_{i - 1}$ is a neighbourhood of $x_{i - 1}$ we can find an open $x_{i - 1} \in W \subset E_{i - 1}$. Since $U_ i$ is dense in $X$ we see that $W \cap U_ i$ is nonempty. Pick any $x_ i \in W \cap U_ i$. By definition of locally quasi-compact spaces we can find a quasi-compact neighbourhood $E_ i$ of $x_ i$ contained in $W \cap U_ i$. Then $E_ i \subset E_{i - 1} \cap U_ i$ as desired. $\square$

Lemma 5.13.4. Let $X$ be a Hausdorff and quasi-compact space. Let $X = \bigcup _{i \in I} U_ i$ be an open covering. Then there exists an open covering $X = \bigcup _{i \in I} V_ i$ such that $\overline{V_ i} \subset U_ i$ for all $i$.

**Proof.**
Let $x \in X$. Choose an $i(x) \in I$ such that $x \in U_{i(x)}$. Since $X \setminus U_{i(x)}$ and $\{ x\} $ are disjoint closed subsets of $X$, by Lemmas 5.12.3 and 5.12.4 there exists an open neighbourhood $U_ x$ of $x$ whose closure is disjoint from $X \setminus U_{i(x)}$. Thus $\overline{U_ x} \subset U_{i(x)}$. Since $X$ is quasi-compact, there is a finite list of points $x_1, \ldots , x_ m$ such that $X = U_{x_1} \cup \ldots \cup U_{x_ m}$. Setting $V_ i = \bigcup _{i = i(x_ j)} U_{x_ j}$ the proof is finished.
$\square$

Lemma 5.13.5. Let $X$ be a Hausdorff and quasi-compact space. Let $X = \bigcup _{i \in I} U_ i$ be an open covering. Suppose given an integer $p \geq 0$ and for every $(p + 1)$-tuple $i_0, \ldots , i_ p$ of $I$ an open covering $U_{i_0} \cap \ldots \cap U_{i_ p} = \bigcup W_{i_0 \ldots i_ p, k}$. Then there exists an open covering $X = \bigcup _{j \in J} V_ j$ and a map $\alpha : J \to I$ such that $\overline{V_ j} \subset U_{\alpha (j)}$ and such that each $V_{j_0} \cap \ldots \cap V_{j_ p}$ is contained in $W_{\alpha (j_0) \ldots \alpha (j_ p), k}$ for some $k$.

**Proof.**
Since $X$ is quasi-compact, there is a reduction to the case where $I$ is finite (details omitted). We prove the result for $I$ finite by induction on $p$. The base case $p = 0$ is immediate by taking a covering as in Lemma 5.13.4 refining the open covering $X = \bigcup W_{i_0, k}$.

Induction step. Assume the lemma proven for $p - 1$. For all $p$-tuples $i'_0, \ldots , i'_{p - 1}$ of $I$ let $U_{i'_0} \cap \ldots \cap U_{i'_{p - 1}} = \bigcup W_{i'_0 \ldots i'_{p - 1}, k}$ be a common refinement of the coverings $U_{i_0} \cap \ldots \cap U_{i_ p} = \bigcup W_{i_0 \ldots i_ p, k}$ for those $(p + 1)$-tuples such that $\{ i'_0, \ldots , i'_{p - 1}\} = \{ i_0, \ldots , i_ p\} $ (equality of sets). (There are finitely many of these as $I$ is finite.) By induction there exists a solution for these opens, say $X = \bigcup V_ j$ and $\alpha : J \to I$. At this point the covering $X = \bigcup _{j \in J} V_ j$ and $\alpha $ satisfy $\overline{V_ j} \subset U_{\alpha (j)}$ and each $V_{j_0} \cap \ldots \cap V_{j_ p}$ is contained in $W_{\alpha (j_0) \ldots \alpha (j_ p), k}$ for some $k$ if there is a repetition in $\alpha (j_0), \ldots , \alpha (j_ p)$. Of course, we may and do assume that $J$ is finite.

Fix $i_0, \ldots , i_ p \in I$ pairwise distinct. Consider $(p + 1)$-tuples $j_0, \ldots , j_ p \in J$ with $i_0 = \alpha (j_0), \ldots , i_ p = \alpha (j_ p)$ such that $V_{j_0} \cap \ldots \cap V_{j_ p}$ is **not** contained in $W_{\alpha (j_0) \ldots \alpha (j_ p), k}$ for any $k$. Let $N$ be the number of such $(p + 1)$-tuples. We will show how to decrease $N$. Since

we find a finite set $K = \{ k_1, \ldots , k_ t\} $ such that the LHS is contained in $\bigcup _{k \in K} W_{i_0 \ldots i_ p, k}$. Then we consider the open covering

The first open on the RHS intersects $V_{j_1} \cap \ldots \cap V_{j_ p}$ in the empty set and the other opens $V_{j_0, k}$ of the RHS satisfy $V_{j_0, k} \cap V_{j_1} \ldots \cap V_{j_ p} \subset W_{\alpha (j_0) \ldots \alpha (j_ p), k}$. Set $J' = J \amalg K$. For $j \in J$ set $V'_ j = V_ j$ if $j \not= j_0$ and set $V'_{j_0} = V_{j_0} \setminus (\overline{V_{j_1}} \cap \ldots \cap \overline{V_{j_ p}})$. For $k \in K$ set $V'_ k = V_{j_0, k}$. Finally, the map $\alpha ' : J' \to I$ is given by $\alpha $ on $J$ and maps every element of $K$ to $i_0$. A simple check shows that $N$ has decreased by one under this replacement. Repeating this procedure $N$ times we arrive at the situation where $N = 0$.

To finish the proof we argue by induction on the number $M$ of $(p + 1)$-tuples $i_0, \ldots , i_ p \in I$ with pairwise distinct entries for which there exists a $(p + 1)$-tuple $j_0, \ldots , j_ p \in J$ with $i_0 = \alpha (j_0), \ldots , i_ p = \alpha (j_ p)$ such that $V_{j_0} \cap \ldots \cap V_{j_ p}$ is **not** contained in $W_{\alpha (j_0) \ldots \alpha (j_ p), k}$ for any $k$. To do this, we claim that the operation performed in the previous paragraph does not increase $M$. This follows formally from the fact that the map $\alpha ' : J' \to I$ factors through a map $\beta : J' \to J$ such that $V'_{j'} \subset V_{\beta (j')}$.
$\square$

Lemma 5.13.6. Let $X$ be a Hausdorff and locally quasi-compact space. Let $Z \subset X$ be a quasi-compact (hence closed) subset. Suppose given an integer $p \geq 0$, a set $I$, for every $i \in I$ an open $U_ i \subset X$, and for every $(p + 1)$-tuple $i_0, \ldots , i_ p$ of $I$ an open $W_{i_0 \ldots i_ p} \subset U_{i_0} \cap \ldots \cap U_{i_ p}$ such that

$Z \subset \bigcup U_ i$, and

for every $i_0, \ldots , i_ p$ we have $W_{i_0 \ldots i_ p} \cap Z = U_{i_0} \cap \ldots \cap U_{i_ p} \cap Z$.

Then there exist opens $V_ i$ of $X$ such that we have $Z \subset \bigcup V_ i$, for all $i$ we have $\overline{V_ i} \subset U_ i$, and we have $V_{i_0} \cap \ldots \cap V_{i_ p} \subset W_{i_0 \ldots i_ p}$ for all $(p + 1)$-tuples $i_0, \ldots , i_ p$.

**Proof.**
Since $Z$ is quasi-compact, there is a reduction to the case where $I$ is finite (details omitted). Because $X$ is locally quasi-compact and $Z$ is quasi-compact, we can find a neighbourhood $Z \subset E$ which is quasi-compact, i.e., $E$ is quasi-compact and contains an open neighbourhood of $Z$ in $X$. If we prove the result after replacing $X$ by $E$, then the result follows. Hence we may assume $X$ is quasi-compact.

We prove the result in case $I$ is finite and $X$ is quasi-compact by induction on $p$. The base case is $p = 0$. In this case we have $X = (X \setminus Z) \cup \bigcup W_ i$. By Lemma 5.13.4 we can find a covering $X = V \cup \bigcup V_ i$ by opens $V_ i \subset W_ i$ and $V \subset X \setminus Z$ with $\overline{V_ i} \subset W_ i$ for all $i$. Then we see that we obtain a solution of the problem posed by the lemma.

Induction step. Assume the lemma proven for $p - 1$. Set $W_{j_0 \ldots j_{p - 1}}$ equal to the intersection of all $W_{i_0 \ldots i_ p}$ with $\{ j_0, \ldots , j_{p - 1}\} = \{ i_0, \ldots , i_ p\} $ (equality of sets). By induction there exists a solution for these opens, say $V_ i \subset U_ i$. It follows from our choice of $W_{j_0 \ldots j_{p - 1}}$ that we have $V_{i_0} \cap \ldots \cap V_{i_ p} \subset W_{i_0 \ldots i_ p}$ for all $(p + 1)$-tuples $i_0, \ldots , i_ p$ where $i_ a = i_ b$ for some $0 \leq a < b \leq p$. Thus we only need to modify our choice of $V_ i$ if $V_{i_0} \cap \ldots \cap V_{i_ p} \not\subset W_{i_0 \ldots i_ p}$ for some $(p + 1)$-tuple $i_0, \ldots , i_ p$ with pairwise distinct elements. In this case we have

is a closed subset of $X$ contained in $U_{i_0} \cap \ldots \cap U_{i_ p}$ not meeting $Z$. Hence we can replace $V_{i_0}$ by $V_{i_0} \setminus T$ to “fix” the problem. After repeating this finitely many times for each of the problem tuples, the lemma is proven. $\square$

Lemma 5.13.7. Let $X$ be a topological space. Let $Z \subset X$ be a quasi-compact subset such that any two points of $Z$ have disjoint open neighbourhoods in $X$. Suppose given an integer $p \geq 0$, a set $I$, for every $i \in I$ an open $U_ i \subset X$, and for every $(p + 1)$-tuple $i_0, \ldots , i_ p$ of $I$ an open $W_{i_0 \ldots i_ p} \subset U_{i_0} \cap \ldots \cap U_{i_ p}$ such that

$Z \subset \bigcup U_ i$, and

for every $i_0, \ldots , i_ p$ we have $W_{i_0 \ldots i_ p} \cap Z = U_{i_0} \cap \ldots \cap U_{i_ p} \cap Z$.

Then there exist opens $V_ i$ of $X$ such that

$Z \subset \bigcup V_ i$,

$V_ i \subset U_ i$ for all $i$,

$\overline{V_ i} \cap Z \subset U_ i$ for all $i$, and

$V_{i_0} \cap \ldots \cap V_{i_ p} \subset W_{i_0 \ldots i_ p}$ for all $(p + 1)$-tuples $i_0, \ldots , i_ p$.

**Proof.**
Since $Z$ is quasi-compact, there is a reduction to the case where $I$ is finite (details omitted). We prove the result in case $I$ is finite by induction on $p$.

The base case is $p = 0$. For $z \in Z \cap U_ i$ and $z' \in Z \setminus U_ i$ there exist disjoint opens $z \in V_{z, z'}$ and $z' \in W_{z, z'}$ of $X$. Since $Z \setminus U_ i$ is quasi-compact (Lemma 5.12.3), we can choose a finite number $z'_1, \ldots , z'_ r$ such that $Z \setminus U_ i \subset W_{z, z'_1} \cup \ldots \cup W_{z, z'_ r}$. Then we see that $V_ z = V_{z, z'_1} \cap \ldots \cap V_{z, z'_ r} \cap U_ i$ is an open neighbourhood of $z$ contained in $U_ i$ with the property that $\overline{V_ z} \cap Z \subset U_ i$. Since $z$ and $i$ where arbitrary and since $Z$ is quasi-compact we can find a finite list $z_1, i_1, \ldots , z_ t, i_ t$ and opens $V_{z_ j} \subset U_{i_ j}$ with $\overline{V_{z_ j}} \cap Z \subset U_{i_ j}$ and $Z \subset \bigcup V_{z_ j}$. Then we can set $V_ i = W_ i \cap (\bigcup _{j : i = i_ j} V_{z_ j})$ to solve the problem for $p = 0$.

Induction step. Assume the lemma proven for $p - 1$. Set $W_{j_0 \ldots j_{p - 1}}$ equal to the intersection of all $W_{i_0 \ldots i_ p}$ with $\{ j_0, \ldots , j_{p - 1}\} = \{ i_0, \ldots , i_ p\} $ (equality of sets). By induction there exists a solution for these opens, say $V_ i \subset U_ i$. It follows from our choice of $W_{j_0 \ldots j_{p - 1}}$ that we have $V_{i_0} \cap \ldots \cap V_{i_ p} \subset W_{i_0 \ldots i_ p}$ for all $(p + 1)$-tuples $i_0, \ldots , i_ p$ where $i_ a = i_ b$ for some $0 \leq a < b \leq p$. Thus we only need to modify our choice of $V_ i$ if $V_{i_0} \cap \ldots \cap V_{i_ p} \not\subset W_{i_0 \ldots i_ p}$ for some $(p + 1)$-tuple $i_0, \ldots , i_ p$ with pairwise distinct elements. In this case we have

is a closed subset of $X$ not meeting $Z$ by our property (3) of the opens $V_ i$. Hence we can replace $V_{i_0}$ by $V_{i_0} \setminus T$ to “fix” the problem. After repeating this finitely many times for each of the problem tuples, the lemma is proven. $\square$

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