Problem Set 1

MAT 127A SQ 2026

Due Friday, April 10th at 11:59PM

Problems Problems

1.

Abbott Example 1.2.5 and Exercise 1.2.6. Look up Example 1.2.5 in the textbook, and then complete Exercise 1.2.6. This is very important for everything, so please work to understand it.

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Verify the triangle inequality in the special case where \(a\) and \(b\) have the same sign.
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Find an efficient proof for all cases at once by first demonstrating \((a+b)^2 \le (|a| + |b|)^2\text{.}\) You can use the fact that \(f(x)=\sqrt{x}\) is increasing: for positive real numbers \(x\) and \(y\text{,}\) \(\sqrt{x} \le \sqrt{y}\) if and only if \(x \le y\text{.}\)
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Prove \(|a-b| \le |a-c| + |c-d| + |d-b|\) for all \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\text{.}\)
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Prove \(||a|-|b|| \le |a-b|\text{.}\)
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Hint.

For part (b), a good general strategy for proving an inequality is to start with the left-hand side and do a sequence of algebraic manipulations or simple inequalities, such as \(a \le |a|\text{,}\) until you arrive at the right-hand side.

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For part (d), remember that \(|x| \le c\) for some positive \(c\in\R\) is the same as \(-c \le x \le c\text{,}\) so you can prove each inequality separately.

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2.

Abbott Exercise 1.3.6. Given sets \(A\) and \(B\text{,}\) define \(A+B = \{a+b : a \in A \text{ and } b \in B\}\text{.}\) Follow these steps to prove that if \(A\) and \(B\) are nonempty and bounded above, then \(\sup(A+B) = \sup A + \sup B\text{.}\)

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Let \(s = \sup A\) and \(t = \sup B\text{.}\) Show \(s+t\) is an upper bound for \(A+B\text{.}\)
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Now let \(u\) be an arbitrary upper bound for \(A+B\text{,}\) and temporarily fix \(a \in A\text{.}\) Show \(t \le u-a\text{.}\)
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Finally, show \(\sup(A+B) = s+t\text{.}\)
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Construct another proof of this same fact using Lemma 1.3.8.
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Hint.

For part (d), to get an element of \(A+B\) within \(\epsilon\) of \(s+t\text{,}\) take elements of \(A\) and \(B\) separately that are sufficiently close to \(s\) and \(t\) and then add them.

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3.

Abbott Exercise 1.3.9.

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If \(\sup A < \sup B\text{,}\) show that there exists an element \(b \in B\) that is an upper bound for \(A\text{.}\)
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Give an example to show that this is not always the case if we only assume \(\sup A \le \sup B\text{.}\)
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4.

Abbott Exercise 1.4.4. Let \(a < b\) be real numbers and consider the set \(T = \Q \cap [a,b]\text{.}\) Show \(\sup T = b\text{.}\) Begin by justifying the fact that \(\sup T\) exists.

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Hint.

Try using Lemma 1.3.8. What about \(\Q\) in particular makes this work?

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5.

Optional. Abbott Exercise 1.3.10. The Cut Property of the real numbers is the following: If \(A\) and \(B\) are nonempty, disjoint sets with \(A \cup B = \R\) and \(a < b\) for all \(a \in A\) and \(b \in B\text{,}\) then there exists \(c \in \R\) such that \(x \le c\) whenever \(x \in A\) and \(x \ge c\) whenever \(x \in B\text{.}\)

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Use the Axiom of Completeness to prove the Cut Property.
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Show that the implication goes the other way; that is, assume \(\R\) possesses the Cut Property and let \(E\) be a nonempty set that is bounded above. Prove \(\sup E\) exists.
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Give a concrete example showing that the Cut Property is not a valid statement when \(\R\) is replaced by \(\Q\text{.}\)
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6.

Optional. Abbott Exercise 1.4.7. Finish the proof of Theorem 1.4.5 by showing that the assumption \(\alpha^2 > 2\) leads to a contradiction of the fact that \(\alpha = \sup T\text{.}\)

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