[proofplan] We prove both directions of the equivalence. For the forward direction, given a locally compact subspace $Y$, we produce an open set $V$ and a closed set $F$ in $X$ with $Y = V \cap F$ by exploiting the Hausdorff property to separate compact neighbourhoods. For the reverse direction, given $Y = V \cap F$ with $V$ open and $F$ closed, we show that local compactness of $X$ transfers to $Y$ by intersecting compact neighbourhoods in $X$ with $F$ to produce compact neighbourhoods in the subspace topology. [/proofplan] [step:Show that a locally compact subspace is locally closed] Assume $Y$ is locally compact in the subspace topology. For each $y \in Y$, there exists an open set $W_y$ in the subspace topology on $Y$ and a compact set $K_y \subset Y$ with $y \in W_y \subset K_y$. Since $W_y$ is open in the subspace topology, there exists an open set $O_y \subset X$ with $W_y = O_y \cap Y$. Define the set \begin{align*} V := \bigcup_{y \in Y} O_y. \end{align*} Since $V$ is a union of open sets in $X$, it is open in $X$, and $Y \subset V$. We claim that $Y$ is closed in the subspace $V$. Let $z \in V \setminus Y$. Then $z \in O_y$ for some $y \in Y$. The set $K_y$ is compact in $Y$ and hence compact in $X$ (compactness is intrinsic). Since $X$ is Hausdorff, $K_y$ is closed in $X$ by the [Compact Subspaces and Hausdorff Spaces](/theorems/307) theorem. Therefore $X \setminus K_y$ is open in $X$. Since $z \notin Y \supset W_y = O_y \cap Y$ and $z \in O_y$, we know $z \notin Y$. In particular $z \notin K_y$, so $z \in X \setminus K_y$. The set $O_y \cap (X \setminus K_y)$ is open in $X$, contains $z$, and is disjoint from $W_y = O_y \cap Y$. Since $W_y \subset K_y$, any point in $O_y \cap (X \setminus K_y)$ lies outside $K_y$ and hence outside $W_y$. Moreover, for any $y' \in Y$ with $y' \in O_y \cap (X \setminus K_y)$, we have $y' \in O_y$ and $y' \in Y$, so $y' \in O_y \cap Y = W_y \subset K_y$, contradicting $y' \notin K_y$. Therefore $O_y \cap (X \setminus K_y)$ is an open neighbourhood of $z$ in $X$ that is disjoint from $Y$, and in particular disjoint from $Y \cap V$. This shows $z \notin \overline{Y}^V$ (the closure of $Y$ in the subspace topology on $V$), so $Y$ is closed in $V$. Setting $F := \overline{Y}$ (the closure in $X$), we have $Y \subset F$ and $Y$ is closed in $V$, so $Y = V \cap \overline{Y}^V$. Since $\overline{Y}^V = \overline{Y} \cap V = F \cap V$, we conclude $Y = V \cap F$, expressing $Y$ as the intersection of the open set $V$ and the closed set $F$ in $X$. [guided] The strategy is to build an open set $V \subset X$ that contains $Y$ and in which $Y$ is closed. Then $Y = V \cap \overline{Y}$, which is the intersection of an open and a closed set. For each $y \in Y$, local compactness of $Y$ in the subspace topology provides an open neighbourhood $W_y$ of $y$ in $Y$ whose closure in $Y$ is compact. Since $Y$ carries the subspace topology, $W_y = O_y \cap Y$ for some open set $O_y \subset X$. Define \begin{align*} V := \bigcup_{y \in Y} O_y. \end{align*} This is open in $X$ and contains $Y$. Now we verify that $Y$ is closed in $V$. Take any $z \in V \setminus Y$. We must find an open neighbourhood of $z$ in $V$ that misses $Y$. Since $z \in V$, we have $z \in O_y$ for some $y \in Y$. The compact set $K_y := \overline{W_y}^Y$ (closure in $Y$) is compact in $Y$ and hence in $X$. Because $X$ is Hausdorff, the [Compact Subspaces and Hausdorff Spaces](/theorems/307) theorem guarantees that $K_y$ is closed in $X$. Since $z \notin Y$, certainly $z \notin K_y$. The set $O_y \cap (X \setminus K_y)$ is open in $X$, contains $z$, and is disjoint from $Y$: any point $y' \in Y \cap O_y \cap (X \setminus K_y)$ would satisfy $y' \in O_y \cap Y = W_y \subset K_y$, contradicting $y' \notin K_y$. So $O_y \cap (X \setminus K_y)$ is an open neighbourhood of $z$ in $V$ that avoids $Y$. This shows $Y$ is closed in $V$, so $Y = V \cap \overline{Y}$ where $V$ is open and $\overline{Y}$ is closed in $X$. [/guided] [/step] [step:Show that a locally closed subspace is locally compact] Assume $Y = V \cap F$ where $V \subset X$ is open and $F \subset X$ is closed. We must show $Y$ is locally compact in the subspace topology. Fix $y \in Y$. Since $X$ is locally compact Hausdorff, there exists an open set $U \subset X$ and a compact set $K \subset X$ with $y \in U \subset K$. Define the open set $U' := U \cap V$, which is open in $X$ and contains $y$. The set $K' := K \cap F$ is the intersection of the compact set $K$ and the closed set $F$ in the Hausdorff space $X$. By the [Compact Subspaces and Hausdorff Spaces](/theorems/307) theorem, $K$ is closed in $X$, so $K \cap F$ is a closed subset of the compact set $K$, and therefore $K'$ is compact. We verify containment: since $U' = U \cap V \subset K \cap V \subset K$ and $Y = V \cap F$, the set $U' \cap Y = U \cap V \cap F$ is open in the subspace topology on $Y$ and satisfies \begin{align*} U' \cap Y = U \cap V \cap F \subset K \cap F = K'. \end{align*} Since $K' \subset F$ and $K' \subset K \subset X$, and $K' \cap V \supset U' \cap Y \ni y$, the set $K' \cap Y = K \cap F \cap V \cap F = K \cap F \cap V = K' \cap V$ is a closed subset of $K'$ (because $Y = V \cap F$ and the closure of $U' \cap Y$ in $Y$ is contained in $K' \cap Y$). In fact, $K' = K \cap F$ is compact, and $K' \cap Y = K \cap F \cap V = K \cap Y$. Since $K' = K \cap F \supset U \cap V \cap F = U' \cap Y$, we have produced, for each $y \in Y$, an open neighbourhood $U' \cap Y$ in the subspace topology whose closure in $Y$ is contained in the compact set $K'$. This closure is a closed subset of the compact set $K'$ and is therefore compact. Hence $Y$ is locally compact. [guided] We are given $Y = V \cap F$ where $V$ is open and $F$ is closed in $X$. Fix $y \in Y$. We need to find an open neighbourhood of $y$ in $Y$ with compact closure in $Y$. Since $X$ is locally compact Hausdorff, there exists an open $U \subset X$ and compact $K \subset X$ with $y \in U \subset K$. The idea is to intersect $U$ with $V$ (to stay inside $Y$'s ambient open set) and intersect $K$ with $F$ (to produce a compact set inside $Y$). Set $U' := U \cap V$, which is open in $X$ and contains $y$. Set $K' := K \cap F$. Since $F$ is closed and $K$ is compact (hence closed in the Hausdorff space $X$ by [Compact Subspaces and Hausdorff Spaces](/theorems/307)), $K'$ is a closed subset of the compact set $K$, hence compact. Now $U' \cap Y = U \cap V \cap (V \cap F) = U \cap V \cap F$. This is open in $Y$. And \begin{align*} U' \cap Y = U \cap V \cap F \subset K \cap F = K'. \end{align*} The closure of $U' \cap Y$ in $Y$ is contained in $\overline{U' \cap Y}^X \cap Y$. Since $U' \cap Y \subset K'$ and $K'$ is closed in $X$, we get $\overline{U' \cap Y}^X \subset K'$. Therefore the closure of $U' \cap Y$ in $Y$ is contained in $K' \cap Y = K \cap F \cap V$, which is a closed subset of $K'$ and hence compact. This gives $y$ an open neighbourhood in $Y$ with compact closure in $Y$, confirming that $Y$ is locally compact. [/guided] [/step] [step:Establish the equivalent formulation with compact closures] The second characterisation follows directly. If $Y$ is locally compact in the subspace topology, then for each $y \in Y$ there exists a compact neighbourhood $K_y$ of $y$ in $Y$, so there exists an open set $W_y$ in $Y$ with $y \in W_y \subset K_y$. Writing $W_y = V_y \cap Y$ for some open $V_y \subset X$, we have $V_y \cap Y \subset K_y$. The closure $\overline{V_y \cap Y}^X$ is a closed subset of $X$ containing $V_y \cap Y$. Since $K_y$ is compact in $Y$ and hence in $X$, and $K_y$ is closed in $X$ (by the Hausdorff property), we have $\overline{V_y \cap Y}^X \subset K_y$, which is compact. Setting $V := V_y$ gives the desired open set. Conversely, if for every $y \in Y$ there exists an open $V \subset X$ with $y \in V$ and $\overline{V \cap Y}^X$ compact, then $V \cap Y$ is open in $Y$ and its closure in $Y$ is $\overline{V \cap Y}^Y = \overline{V \cap Y}^X \cap Y$, which is a closed subset of the compact set $\overline{V \cap Y}^X$ and hence compact. This gives a compact neighbourhood of $y$ in $Y$. [/step]