Question

Let \( C[0,1] = \{ f: [0,1] \to \mathbb{R} : f \text{ is continuous}\} \) and \[ d_\infty(f,g) = \sup\{ |f(x) - g(x)|: x \in [0, 1]\} \] for \( f, g \in C[0,1] \).
For each \( n \in \mathbb{N} \), define \( f_n: [0,1] \to \mathbb{R} \) by \( f_n(x) = x^n \) for all \( x \in [0,1] \).
Let \( P = \{f_n: n \in \mathbb{N}\} \).
Which of the following statements is/are correct?

• A. \( P \) is totally bounded in \( (C[0, 1], d_\infty) \)
• B. \( P \) is bounded in \( (C[0, 1], d_\infty) \)
• C. \( P \) is closed in \( (C[0, 1], d_\infty) \)
• D. \( P \) is open in \( (C[0, 1], d_\infty) \)

Hint:

For questions involving properties of function sets in \( (C[a,b], d_\infty) \), the Arzelà-Ascoli theorem is a powerful tool. Remember its conditions: pointwise boundedness and equicontinuity. If a set fails to be equicontinuous, it cannot be totally bounded (or compact).

Answer 1

The Correct Option is B, C

Step 1: Understanding the Concept:
This question deals with properties of a set of functions \( P = \{x^n\} \) in the metric space of continuous functions on \( [0,1] \) with the supremum norm (\( d_\infty \)). We need to determine if the set \( P \) is bounded, closed, open, or totally bounded.
Step 2: Key Formula or Approach:
- Bounded: A set \( S \) is bounded if there exists a function \( g \) and a radius \( M>0 \) such that \( d_\infty(f, g) \leq M \) for all \( f \in S \). It's sufficient to check the distance from the zero function.
- Closed: A set \( S \) is closed if it contains all its limit points. An equivalent statement is that for any sequence in \( S \) that converges in the space, its limit is also in \( S \).
- Open: A set \( S \) is open if for every point \( f \in S \), there exists an \( \epsilon \)-ball centered at \( f \) that is entirely contained in \( S \).
- Totally Bounded: A set \( S \) is totally bounded if for any \( \epsilon>0 \), \( S \) can be covered by a finite number of open balls of radius \( \epsilon \). In \( C[0,1] \), this is related to the Arzelà-Ascoli theorem.
Step 3: Detailed Explanation or Calculation:
Analysis of Statement (B): Boundedness
Let's check the distance of any element \( f_n \in P \) from the zero function \( g(x)=0 \).
\[ d_\infty(f_n, 0) = \sup_{x \in [0,1]} |f_n(x) - 0| = \sup_{x \in [0,1]} |x^n| \] Since \( x \in [0,1] \), we have \( 0 \leq x^n \leq 1 \). The supremum is achieved at \( x=1 \), where \( x^n=1 \).
So, \( d_\infty(f_n, 0) = 1 \) for all \( n \in \mathbb{N} \).
Since the distance of every function in \( P \) from the zero function is 1, the set \( P \) is contained within the ball of radius 1 (or any radius \( M \geq 1 \)) centered at the zero function.
Therefore, statement (B) is correct.
Analysis of Statement (C) and (D): Closed/Open
A set is closed if it contains all its limit points. Let's consider a sequence from \( P \), say \( \{f_{n_k}\} \), and assume it converges to a function \( g \in C[0,1] \) with respect to the \( d_\infty \) metric. Convergence in \( d_\infty \) is uniform convergence.
The sequence of functions \( \{f_n(x) = x^n\} \) converges pointwise to the function:
\[ f(x) = \begin{cases} 0, & 0 \le x<1
1, & x=1 \end{cases} \] This limit function \( f(x) \) is not continuous, so it is not in \( C[0,1] \). Therefore, the sequence \( \{f_n\} \) cannot converge uniformly on \( [0,1] \). This means that no subsequence of \( \{f_n\} \) can converge to a function in \( C[0,1] \). Thus, the set \( P \) has no limit points in \( C[0,1] \). A set with no limit points is, by definition, closed.
Therefore, statement (C) is correct.
A set is open if it contains an open ball around each of its points. Let \( f_m \in P \). Consider any \( \epsilon>0 \). The function \( g(x) = x^m + \epsilon/2 \) is in \( C[0,1] \) and \( d_\infty(g, f_m) = \epsilon/2<\epsilon \). However, \( g(x) \) is not of the form \( x^n \) for any \( n \in \mathbb{N} \), so \( g \notin P \). This means no open ball around \( f_m \) is fully contained in \( P \).
Therefore, statement (D) is incorrect.
Analysis of Statement (A): Totally Bounded
A subset of a complete metric space (like \( C[0,1] \)) is compact if and only if it is closed and totally bounded. We have already established that \( P \) is closed. If \( P \) were totally bounded, it would be compact. By the Arzelà-Ascoli theorem, a set of functions in \( C[0,1] \) is relatively compact (its closure is compact) if and only if it is pointwise bounded and equicontinuous.
The set \( P \) is pointwise bounded: \( |f_n(x)| = |x^n| \leq 1 \) for all \( x \in [0,1] \) and all \( n \).
Let's check for equicontinuity. The family \( P \) is equicontinuous if for every \( \epsilon>0 \), there exists \( \delta>0 \) such that for all \( n \in \mathbb{N} \) and all \( x,y \in [0,1] \) with \( |x-y|<\delta \), we have \( |f_n(x) - f_n(y)|<\epsilon \).
Let's choose \( \epsilon = 1/2 \). For any \( \delta>0 \), let's pick \( y=1 \) and \( x = 1 - \delta/2 \). Then \( |x-y| = \delta/2<\delta \). Consider \( |f_n(1) - f_n(1-\delta/2)| = |1^n - (1-\delta/2)^n| = 1 - (1-\delta/2)^n \).
As \( n \to \infty \), \( (1-\delta/2)^n \to 0 \). So, we can choose \( n \) large enough such that \( 1 - (1-\delta/2)^n>1/2 = \epsilon \). Since we cannot find a single \( \delta \) that works for all \( n \), the set \( P \) is not equicontinuous.
By Arzelà-Ascoli, the closure of \( P \) is not compact. Since \( P \) is closed, \( P \) itself is not compact. Since \( P \) is closed but not compact, it cannot be totally bounded. Therefore, statement (A) is incorrect.
Step 4: Final Answer:
The correct statements are (B) and (C).
Step 5: Why This is Correct:
The set \( P \) is a collection of distinct functions \( x, x^2, x^3, ...... \). It's bounded because all functions stay within the range [0, 1]. It's closed because the sequence does not converge to any function *within* the space \( C[0,1] \), meaning it has no limit points to fail to contain. It's not totally bounded because the functions become arbitrarily steep near \( x=1 \) for large \( n \), violating the equicontinuity condition required for compactness in function spaces.