Exercise · ZW-HEINE-BOREL-THEOREM-004 · ★☆☆ Basic
Why is \((0,1]\) not compact in \(\mathbb{R}\)?
Choose one option.
Show answer
- B. It is bounded but not closed
Explanation
It is bounded, but its complement \((-\infty,0]\cup(1,\infty)\) is not open, so it is not closed. Being infinite does not prevent compactness.
Review the concept: Heine-Borel theorem
A subset \(K\) of \(\mathbb{R}^n\) with the usual metric is compact if and only if it is closed and bounded.
\[K\subseteq\mathbb{R}^n\ \text{compact}\iff K\ \text{closed and bounded}\]
More practice problems: heine-borel theorem
- ZW-HEINE-BOREL-THEOREM-001 Identify the compact subset of ℝ:
- ZW-HEINE-BOREL-THEOREM-002 Which of these subsets of ℝ^2 is compact?
- ZW-HEINE-BOREL-THEOREM-003 Why is [0,∞) not compact in ℝ?
- ZW-HEINE-BOREL-THEOREM-005 Among these plane figures, which one is compact?
- ZW-HEINE-BOREL-THEOREM-006 Which of these subsets of ℝ is compact?
- ZW-HEINE-BOREL-THEOREM-007 Which of these subsets of ℝ^3 is compact?
- ZW-HEINE-BOREL-THEOREM-008 Let A=[0,1], B=(0,1), C=ℤ, D=\0\∪\1/n:n≥1\ in ℝ. How many are compact?
- ZW-HEINE-BOREL-THEOREM-009 The metric space X=(0,1) with the usual distance is closed (in itself) and bounded. Is it compact?