Let \(X\) be a noetherian topological space and let \(\mcp\) be a property of closed subsets of \(X\). Assume that for any closed subset \(Y\) of \(X\), if \(\mcp\) holds for every proper closed subset of \(Y\) then \(\mcp\) holds for \(Y\). In particular, \(\mcp\) must hold for the empty set. Then \(\mcp\) holds for \(X\).
Solution
1. In a noetherian topological space, every nonempty collection of closed subsets has a minimal element under inclusion.
A topological space is noetherian exactly when every descending chain of closed subsets stabilizes.
Let \(\mathcal C\) be a nonempty collection of closed subsets. If it had no minimal element, choose \[ Y_0\in\mathcal C. \] Since \(Y_0\) is not minimal, choose \[ Y_1\in\mathcal C, \qquad Y_1\subsetneq Y_0. \] Inductively one obtains a strictly descending chain \[ Y_0\supsetneq Y_1\supsetneq Y_2\supsetneq\cdots, \] contradicting noetherianity. Hence a minimal member exists.
2. Suppose, for contradiction, that \(\mcp\) does not hold for every closed subset of \(X\).Let\[ \mathcal B = \{Y\subseteq X:Y\text{ closed and }\mcp(Y)\text{ fails}\}. \]Choose a minimal member\[ Y\in\mathcal B. \]
Under the contradictory assumption, \(\mathcal B\) is nonempty. Step 1 therefore supplies a minimal member.
3. Every proper closed subset\[ Z\subsetneq Y \]satisfies \(\mcp\).
If some proper closed \(Z\subsetneq Y\) failed \(\mcp\), then \(Z\in\mathcal B\), contradicting minimality of \(Y\) in \(\mathcal B\).
4. The property \(\mcp\) holds for \(Y\), a contradiction.
By 3, \(\mcp\) holds for every proper closed subset of \(Y\). The induction hypothesis stated in the problem therefore implies that \(\mcp\) holds for \(Y\) itself. This contradicts \(Y\in\mathcal B\).
5. Hence \(\mcp\) holds for every closed subset of \(X\), in particular for \(X\).
Step 4 shows that the family \(\mathcal B\) of counterexamples must be empty. Since \(X\) is a closed subset of itself, \(\mcp(X)\) holds.
6. Q.E.D.
Step 5 is the desired conclusion.