Problem A5: Accumulation Points and Closed Sets

(a) Definition of Accumulation Point

A point \(a \in \mathbb{R}^n\) is an accumulation point of a set \(A \subseteq \mathbb{R}^n\) if every neighborhood of \(a\) contains infinitely many points of \(A\). Formally:

\[ \forall \epsilon > 0, \quad \exists \, x \in A \text{ such that } x \neq a \text{ and } \|x - a\| < \epsilon. \]

(b) Proof: Closed Sets Contain All Accumulation Points

Theorem: \(A\) is closed \(\iff\) \(A\) contains all its accumulation points.

Proof:

  1. (\(\Rightarrow\)) Assume \(A\) is closed. Let \(a\) be an accumulation point of \(A\). Suppose for contradiction \(a \notin A\). Then \(a \in A^c\), which is open. Thus, \(\exists \epsilon > 0\) such that \(B(a,\epsilon) \subseteq A^c\), contradicting \(a\) being an accumulation point.

  2. (\(\Leftarrow\)) Assume \(A\) contains all its accumulation points. To show \(A^c\) is open: Let \(x \in A^c\). Since \(x\) is not an accumulation point, \(\exists \epsilon > 0\) such that \(B(x,\epsilon) \cap A = \emptyset\). Hence, \(B(x,\epsilon) \subseteq A^c\), so \(A^c\) is open.

(c) Sequence Characterization

Theorem: \(x\) is an accumulation point of \(A\) \(\iff\) \(\exists \{x_n\} \subseteq A\) with \(x_n \neq x\) \(\forall n\) and \(x_n \to x\).

Proof:

  1. (\(\Rightarrow\)) For each \(n \in \mathbb{N}\), pick \(x_n \in B(x,1/n) \cap A\) with \(x_n \neq x\). Then \(\|x_n - x\| < 1/n \to 0\).

  2. (\(\Leftarrow\)) Given \(x_n \to x\) and \(x_n \neq x\), any neighborhood of \(x\) contains infinitely many \(x_n\), so \(x\) is an accumulation point.

(d) Closure and Boundary

Theorem: \(\overline{A} = A \cup A' = A \cup \partial A\), where: - \(A'\) = set of accumulation points - \(\partial A\) = boundary points (every neighborhood intersects both \(A\) and \(A^c\))

Proof: 1. Clearly \(A \cup A' \subseteq \overline{A}\). 2. If \(x \in \overline{A} \setminus A\), every neighborhood of \(x\) intersects \(A\), so \(x \in A'\). 3. For \(\partial A\): Note \(\overline{A} = A \cup \partial A\) since \(\partial A\) includes limit points not in \(A\).

Problem A8: Epsilon-Delta Proof

Statement

Prove using the \(\epsilon\)-\(\delta\) definition: \[ \lim_{(x,y) \to (-1,2)} (3x - 4y + 5) = -6. \]

solution

  1. Goal: For any \(\epsilon > 0\), find \(\delta > 0\) such that: \[ \sqrt{(x + 1)^2 + (y - 2)^2} < \delta \implies |3x - 4y + 11| < \epsilon. \]

  2. Rewrite the expression: \[ |3x - 4y + 11| = |3(x + 1) - 4(y - 2)| \leq 3|x + 1| + 4|y - 2|. \]

  3. Bound the terms: Since \(|x + 1| \leq \sqrt{(x + 1)^2 + (y - 2)^2} < \delta\) and similarly for \(|y - 2|\): \[ 3|x + 1| + 4|y - 2| < 3\delta + 4\delta = 7\delta. \]

  4. Choose \(\delta\): Set \(7\delta \leq \epsilon\), i.e., \(\delta = \frac{\epsilon}{7}\).

  5. Verification: If \(\|(x,y) - (-1,2)\| < \delta\), then: \[ |3x - 4y + 11| < 7\delta = 7 \left( \frac{\epsilon}{7} \right) = \epsilon. \quad \blacksquare \]