Let \(X,Y\) be metric spaces and let \(f:X\to Y\) be continuous. Let \(K_1,K_2,\ldots\) be nonempty compact subsets of \(X\) with \[ K_{n+1}\subseteq K_n \] for all \(n\), and put \[ K=\bigcap_{n=1}^{\infty}K_n. \] Prove that \[ f(K)=\bigcap_{n=1}^{\infty}f(K_n). \]
Solution
1. One has\[ f(K)\subseteq\bigcap_{n=1}^{\infty}f(K_n). \]
Since \[ K\subseteq K_n \] for every \(n\), one has \[ f(K)\subseteq f(K_n) \] for every \(n\). Taking the intersection over \(n\) gives the inclusion.
2. Let\[ y\in\bigcap_{n=1}^{\infty}f(K_n). \]For each \(n\), choose \(x_n\in K_n\) such that\[ f(x_n)=y. \]
The membership \(y\in f(K_n)\) gives at least one such \(x_n\) for every \(n\).
3. The sequence \((x_n)\) has a subsequence converging to some point \(x\in K_1\).
Nestedness gives \(K_n\subseteq K_1\) for every \(n\), so the entire sequence lies in the compact metric space \(K_1\). Sequential compactness yields a convergent subsequence \[ x_{n_j}\longrightarrow x\in K_1. \]
4. The limit point \(x\) belongs to every \(K_m\).
Fix \(m\). For all sufficiently large \(j\), one has \(n_j\ge m\), and therefore \[ x_{n_j}\in K_{n_j}\subseteq K_m. \] The compact set \(K_m\) is closed in the metric space \(X\), so the limit of this tail also belongs to \(K_m\). Thus \(x\in K_m\). Since \(m\) was arbitrary, \[ x\in\bigcap_{m=1}^{\infty}K_m=K. \]
5. One has \(f(x)=y\).
By continuity of \(f\) and step 3, \[ f(x) = \lim_{j\to\infty}f(x_{n_j}) = \lim_{j\to\infty}y =y. \]
6. Therefore\[ \bigcap_{n=1}^{\infty}f(K_n)\subseteq f(K). \]
Starting with an arbitrary \(y\) in the left-hand side, steps <1>2–<1>5 produce \(x\in K\) with \(f(x)=y\).
7. Hence\[ \boxed{f(K)=\bigcap_{n=1}^{\infty}f(K_n)}. \]
Combine steps 1 and 6.
8. Q.E.D.
Step 7 is the required equality.