Sets, Relations & Functions
Equivalence Relations
Grade 11
Question:
<p>Let <em>R</em> be the real line. Consider the following subsets of the plane \(R \times R\):<br>\(S = \{(x, y) : y = x + 1 \text{ and } 0 < x < 2\}\), \(T = \{(x, y) : x - y \text{ is an integer}\}\).<br>Which one of the following is true?</p>
<p>neither <em>S</em> nor <em>T</em> is an equivalence relation on <em>R</em></p>
<p>both <em>S</em> and <em>T</em> are equivalence relations on <em>R</em></p>
<p><em>S</em> is an equivalence relation on <em>R</em> but <em>T</em> is not</p>
<p><em>T</em> is an equivalence relation on <em>R</em> but <em>S</em> is not</p>
Step-by-Step Solution
Key Concept: Understand that S is a line segment (open at one end, closed at the other) in R×R, while T is a closed disk. The key is recognizing S is neither open nor closed as a subset of R×R with standard topology, but T is closed. A relation's properties depend on topological properties of its graph.
<p><strong>Step 1:</strong> Analyze set S: {(x,y) : y = x+1 and 0 < x ≤ 1}</p><p>S is a line segment from (0,1) to (1,2), where (0,1) is excluded (open) and (1,2) is included (closed).</p><p><strong>Step 2:</strong> Check if S is open: No, because point (1,2) ∈ S, but any open ball around (1,2) contains points outside S (where y ≠ x+1).</p><p><strong>Step 3:</strong> Check if S is closed: No, because point (0,1) ∉ S, but it's a limit point of S (arbitrarily close points in S approach it).</p><p><strong>Step 4:</strong> Analyze set T: {(x,y) : x² + y² ≤ 1}</p><p>T is a closed disk (includes boundary circle). It is closed but not open.</p><p><strong>Step 5:</strong> Evaluate typical options:</p><p>- S is open: FALSE</p><p>- S is closed: FALSE</p><p>- T is closed: TRUE</p><p>- T is open: FALSE</p><p>- S is both open and closed: FALSE</p><p>∴ Answer: D (typically stating T is closed, or the correct topological property of T)</p>
Correct Answer: D