Consider the following subset of the metric space \((C_b([0,1]),\rho_\infty)\): \[A:=\{f\in C_b([0,1]):f([0,1])\subseteq[0,1]\}.\]
Determine whether \(A\) is bounded, and if so what is its diameter.
Determine whether \(A\) is closed in \(C_b([0,1])\).
Determine whether \(A\) is compact in \(C_b([0,1])\).
Solution
Goal: For \(A = \{f \in C_b([0,1]) : f([0,1]) \subseteq [0,1]\}\) in \((C_b[0,1], \rho_\infty)\): (a) bounded? diameter? (b) closed? (c) compact?
1. (a) \(A\) is bounded, with diameter \(1\).
1.1. \(\rho_\infty(f,g) \le 1\) for all \(f, g \in A\).Proof: \(f(x), g(x) \in [0,1]\) for all \(x\), so \(|f(x) - g(x)| \le 1\).
1.2. The diameter is exactly \(1\).Proof: the constant functions \(0\) and \(1\) are in \(A\) and \(\rho_\infty(0, 1) = 1\).
1.3. Q.E.D. Proof: 1 and
1.2.
2. (b) \(A\) is closed in \(C_b[0,1]\).
2.1. Let \(f_n \in A\) with \(f_n \to f\) uniformly (in \(\rho_\infty\)); show \(f \in A\).
2.2. \(f(x) \in [0,1]\) for every \(x\).Proof: \(f(x) = \lim_n f_n(x)\) and each \(f_n(x) \in [0,1]\), a closed set.
2.3. Q.E.D. Proof: 2 shows \(f([0,1]) \subseteq [0,1]\), so \(f \in A\); \(A\) is closed.
3. (c) \(A\) is NOT compact.
3.1. \(A\) is closed (by 2) and bounded (by <1>1), but not equicontinuous.
3.2. The sequence \(f_n(x) = x^n\) lies in \(A\) and has no uniformly convergent subsequence.Proof: each \(f_n\) maps \([0,1]\) into \([0,1]\); any subsequence converges pointwise to the discontinuous limit \(f(x) = 0\) on \([0,1)\), \(f(1) = 1\), so no subsequence converges uniformly (a uniform limit of continuous functions is continuous).
3.3. Arzelà–Ascoli: a subset of \(C[0,1]\) is compact iff closed, bounded, and equicontinuous; \(A\) fails equicontinuity.Proof: 2 exhibits a sequence without a convergent subsequence, so \(A\) is not sequentially compact, hence not compact.
3.4. Q.E.D. Proof: <2>2–<2>3 show \(A\) is not compact.