Analysis III: Week-1

Definition (Continuity at a point, \( \epsilon\text{–}\delta \))

Let \( I = (a,b) \subseteq \mathbb{R} \), and let \( f \) be a function on \( I \). Then \( f \) is said to be continuous at \( x \in I \) if for a given \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( |x-y| < \delta \), it implies \( |f(x)-f(y)| < \epsilon \).

This can be written as:

If \( y \in (x-\delta, x+\delta) \), then \( f(y) \in (f(x)-\epsilon, f(x)+\epsilon) \).

The notes also relate this to the inverse image:

\( f^{-1}((f(x)-\epsilon, f(x)+\epsilon)) = \{y \in I \mid f(y) \in (f(x)-\epsilon, f(x)+\epsilon) \} \)

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Definition (Open Set)

A subset \( U \subseteq \mathbb{R} \) is called open if either \( U = \emptyset \) or for any given \( x \in U \), there exists an \( \epsilon > 0 \) such that \( (x-\epsilon, x+\epsilon) \subseteq U \).

Properties:

• \( \emptyset \) and \( \mathbb{R} \) are open.
• If \( U_1, U_2 \) are open subsets of \( \mathbb{R} \), then their intersection \( U_1 \cap U_2 \) is open.
• If \( \{U_\alpha\}_{\alpha \in J} \) is a collection of open sets, then their union \( \bigcup_{\alpha \in J} U_\alpha \) is open.

Exercise:

• Every open interval is an open set.
• Closed intervals \([\, , \,]\), half-open intervals \([\, , \,) \) or \( (\, , \,] \) are not open subsets of \( \mathbb{R} \).

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Equivalent Definition (Continuity at a point, Topological)

Let \( f: \mathbb{R} \to \mathbb{R} \) and \( x \in \mathbb{R} \). The function \( f \) is said to be continuous at \( x \) if for any open set \( U \subseteq \mathbb{R} \) such that \( f(x) \in U \), there exists an open set \( V \subseteq \mathbb{R} \) such that \( x \in V \) and \( f(V) \subseteq U \).

Continuation of Equivalence Proof

If \( f(x) \in U \), where \( U \) is an open set, then \( \exists \epsilon > 0 \) such that \( (f(x)-\epsilon, f(x)+\epsilon) \subseteq U \).

From the \( \epsilon\text{–}\delta \) definition of continuity at \( x \), we know \( \exists \delta > 0 \) such that \( (x-\delta, x+\delta) \subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon)) \subseteq f^{-1}(U) \).

Definition of a Continuous Function

A function \( f: I \to \mathbb{R} \) is continuous if it is continuous at every point of its domain.

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Proposition

Proposition: A function \( f: \mathbb{R} \to \mathbb{R} \) is continuous if and only if for every open set \( U \subseteq \mathbb{R} \), the inverse image \( f^{-1}(U) \subseteq \mathbb{R} \) is also open.

Proof:

\( (\Rightarrow) \) Forward Direction:

Assume \( f \) is continuous. Let \( U \subseteq \mathbb{R} \) be an open set. We need to show that \( f^{-1}(U) \) is open.

Let \( x \in f^{-1}(U) \). By definition, this means \( f(x) \in U \). Since \( U \) is an open set, there exists an \( \epsilon > 0 \) such that the open interval \( (f(x)-\epsilon, f(x)+\epsilon) \subseteq U \).

Because \( f \) is continuous at \( x \), for this specific \( \epsilon \), there must exist a \( \delta > 0 \) such that the interval \( (x-\delta, x+\delta) \) is mapped by \( f \) into \( (f(x)-\epsilon, f(x)+\epsilon) \).

Therefore, \( (x-\delta, x+\delta) \subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon)) \subseteq f^{-1}(U) \).

Since for any arbitrary \( x \in f^{-1}(U) \), we found an open interval around \( x \) that is fully contained in \( f^{-1}(U) \), the set \( f^{-1}(U) \) is open by definition.

\( (\Leftarrow) \) Backward Direction:

Assume that for every open set \( U \), its inverse image \( f^{-1}(U) \) is open. We need to show that \( f \) is continuous at every point \( x \).

Let \( x \in \mathbb{R} \) be an arbitrary point, and let \( \epsilon > 0 \) be given. Consider the interval \( U = (f(x)-\epsilon, f(x)+\epsilon) \). This is an open set in \( \mathbb{R} \).

By our assumption, the inverse image \( f^{-1}(U) \) must be an open set. We know that \( x \in f^{-1}(U) \) because \( f(x) \in U \).

Since \( f^{-1}(U) \) is open and contains \( x \), by the definition of an open set, there must exist a \( \delta > 0 \) such that the open interval \( (x-\delta, x+\delta) \subseteq f^{-1}(U) \).

This implies that for any \( y \in (x-\delta, x+\delta) \), we have \( f(y) \in U \), which means \( |f(y)-f(x)| < \epsilon \).

This is precisely the \( \epsilon\text{–}\delta \) definition of continuity at \( x \). Thus, \( f \) is continuous.

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Definitions in \( \mathbb{R}^n \)

Neighborhood in \( \mathbb{R} \):

For \( x \in \mathbb{R} \) and \( \epsilon > 0 \):

• The open \( \epsilon \)-neighborhood is \( N_\epsilon(x) = \{ a \in \mathbb{R} \mid |x-a| < \epsilon \} \).
• The closed \( \epsilon \)-neighborhood is \( \bar{N}_\epsilon(x) = \{ a \in \mathbb{R} \mid |x-a| \le \epsilon \} \).

Euclidean Norm in \( \mathbb{R}^n \):

For a vector \( x = (x_1, \dots, x_n) \in \mathbb{R}^n \), the norm (or length) is:

\[ |x| = \sqrt{x_1^2 + \dots + x_n^2} \]

Neighborhood in \( \mathbb{R}^n \) (Ball):

For a point \( x \in \mathbb{R}^n \) and \( \epsilon > 0 \):

• The open ball is \( N_\epsilon(x) = \{ a \in \mathbb{R}^n \mid |x-a| < \epsilon \} \).
• The closed ball is \( \bar{N}_\epsilon(x) = \{ a \in \mathbb{R}^n \mid |x-a| \le \epsilon \} \).