Let p ∶ \(\mathbb{R}\)2 → \(\mathbb{R}\) be defined by \(p(x, y)=\left\{\begin{array}{lll} |x| & \text { if } & x \neq 0 \\ |y| & \text { if } & x=0 . \end{array}\right.\) Which of the following statements are true?
Function Definition
The function \(p\) is defined on \(\mathbb{R}^2\) and maps to \(\mathbb{R}\). Its definition is given as:
\[ p(x, y) = \left\{ \begin{array}{lll} |x| & \text{ if } & x \neq 0 \\ |y| & \text{ if } & x=0 \end{array} \right. \]
We will analyze each statement about this function \(p(x, y)\) to determine which ones are true.
Statement 1: p(x, y) = 0 Condition
The first statement claims that \(p(x, y) = 0\) if and only if \(x = y = 0\).
- Let's check the "if" part: If \(x = y = 0\), is \(p(x, y) = 0\)? When \(x = 0\), the definition of \(p(x, y)\) uses the second case, which is \(p(x, y) = |y|\). Substituting \(x=0\) and \(y=0\) into the definition, we get \(p(0, 0) = |0|\). The absolute value of 0 is 0, so \(p(0, 0) = 0\). Thus, if \(x = y = 0\), then \(p(x, y) = 0\). This part is true.
- Let's check the "only if" part: If \(p(x, y) = 0\), is \(x = y = 0\)?
Assume \(p(x, y) = 0\). We look at the two cases in the definition of \(p(x, y)\):
- Case A: \(x \neq 0\). In this case, \(p(x, y) = |x|\). If \(p(x, y) = 0\), then \(|x| = 0\). The only way for the absolute value of a real number to be 0 is if the number itself is 0, so \(x = 0\). This contradicts our assumption for this case that \(x \neq 0\). So, this case is not possible if \(p(x, y) = 0\).
- Case B: \(x = 0\). In this case, \(p(x, y) = |y|\). If \(p(x, y) = 0\), then \(|y| = 0\). This implies \(y = 0\). In this case, we have \(x = 0\) and \(y = 0\).
Since both directions hold, the statement "\(p(x, y) = 0\) if and only if \(x = y = 0\)" is true.
Statement 2: p(x, y) Non-Negativity
The second statement says \(p(x, y) \ge 0\) for all \(x, y \in \mathbb{R}^2\).
Let's examine the definition of \(p(x, y)\) in both cases:
- If \(x \neq 0\), then \(p(x, y) = |x|\). The absolute value of any real number \(x\) is always greater than or equal to 0 (i.e., \(|x| \ge 0\)). Thus, \(p(x, y) \ge 0\) in this case.
- If \(x = 0\), then \(p(x, y) = |y|\). Similarly, the absolute value of any real number \(y\) is always greater than or equal to 0 (i.e., \(|y| \ge 0\)). Thus, \(p(x, y) \ge 0\) in this case.
In every case defined, the value of \(p(x, y)\) is an absolute value, which means it is always non-negative.
Thus, the statement "\(p(x, y) \ge 0\) for all \(x, y\)" is true.
Statement 3: p(αx, αy) Homogeneity
The third statement asserts that \(p(\alpha x, \alpha y) = |\alpha| p(x, y)\) for all \(\alpha \in \mathbb{R}\) and for all \(x, y \in \mathbb{R}^2\). This is the property of homogeneity with respect to scaling by a scalar \(\alpha\).
Let's check this property by considering different situations:
- Consider the case where \(x \neq 0\). In this case, \(p(x, y) = |x|\).
- If \(\alpha \neq 0\), then \(\alpha x \neq 0\). According to the definition of \(p\), \(p(\alpha x, \alpha y) = |\alpha x|\). We know that \(|\alpha x| = |\alpha||x|\). Substituting \(|x| = p(x, y)\), we get \(p(\alpha x, \alpha y) = |\alpha| p(x, y)\). This holds.
- If \(\alpha = 0\), then \(\alpha x = 0 \cdot x = 0\). According to the definition, \(p(\alpha x, \alpha y) = p(0, 0 \cdot y) = p(0, 0)\). Since the first component is 0, \(p(0, 0) = |0| = 0\). Now consider the right side: \(|\alpha| p(x, y) = |0| p(x, y) = 0 \cdot p(x, y)\). Since \(x \neq 0\), \(p(x, y) = |x|\). So the right side is \(0 \cdot |x| = 0\). Both sides equal 0, so \(p(\alpha x, \alpha y) = |\alpha| p(x, y)\) holds when \(\alpha = 0\) and \(x \neq 0\).
- Consider the case where \(x = 0\). In this case, \(p(x, y) = |y|\).
- If \(\alpha \neq 0\), then \(\alpha x = \alpha \cdot 0 = 0\). According to the definition, \(p(\alpha x, \alpha y) = p(0, \alpha y)\). Since the first component is 0, \(p(0, \alpha y) = |\alpha y|\). We know that \(|\alpha y| = |\alpha||y|\). Substituting \(|y| = p(x, y)\) (when \(x=0\)), we get \(p(\alpha x, \alpha y) = |\alpha| p(x, y)\). This holds.
- If \(\alpha = 0\), then \(\alpha x = 0\) and \(\alpha y = 0\). According to the definition, \(p(\alpha x, \alpha y) = p(0, 0) = |0| = 0\). Now consider the right side: \(|\alpha| p(x, y) = |0| p(0, y) = 0 \cdot |y| = 0\). Both sides equal 0, so \(p(\alpha x, \alpha y) = |\alpha| p(x, y)\) holds when \(\alpha = 0\) and \(x = 0\).
In every possible situation, the property \(p(\alpha x, \alpha y) = |\alpha| p(x, y)\) is satisfied.
Thus, the statement "\(p(\alpha x, \alpha y) = |\alpha| p(x, y)\) for all \(\alpha \in \mathbb{R}\) and for all \(x, y\)" is true.
Statement 4: p(x1 + x2, y1 + y2) Triangle Inequality
The fourth statement tests the triangle inequality: \(p(x_1 + x_2, y_1 + y_2) \le p(x_1, y_1) + p(x_2, y_2)\) for all \((x_1, y_1), (x_2, y_2) \in \mathbb{R}^2\).
Let's try to find specific points \((x_1, y_1)\) and \((x_2, y_2)\) to see if this inequality holds. If we find even one pair of points where it fails, the statement is false.
Consider the points \((x_1, y_1) = (1, 5)\) and \((x_2, y_2) = (-1, 5)\).
- Calculate \(p(x_1, y_1)\): Here, \(x_1 = 1\), which is not 0. So, \(p(1, 5) = |x_1| = |1| = 1\).
- Calculate \(p(x_2, y_2)\): Here, \(x_2 = -1\), which is not 0. So, \(p(-1, 5) = |x_2| = |-1| = 1\).
- The sum of \(p(x_1, y_1)\) and \(p(x_2, y_2)\) is \(1 + 1 = 2\).
Now, let's calculate the left side of the inequality, \(p(x_1 + x_2, y_1 + y_2)\).
- First, add the vectors: \((x_1 + x_2, y_1 + y_2) = (1 + (-1), 5 + 5) = (0, 10)\).
- Next, calculate \(p(0, 10)\). Here, the first component is 0. According to the definition of \(p\), when the first component is 0, we use the absolute value of the second component. So, \(p(0, 10) = |10| = 10\).
Now, let's compare the left side and the right side using the inequality: \(p(x_1 + x_2, y_1 + y_2) \le p(x_1, y_1) + p(x_2, y_2)\) Substituting the calculated values: \(10 \le 1 + 1\) \(10 \le 2\)
The statement \(10 \le 2\) is false.
Since we found a specific pair of points \((1, 5)\) and \((-1, 5)\) for which the triangle inequality does not hold, the statement "\(p(x_1 + x_2, y_1 + y_2) \le p(x_1, y_1) + p(x_2, y_2)\) for all \((x_1, y_1), (x_2, y_2)\)" is false.
Conclusion
Based on our analysis:
- Statement 1 is true.
- Statement 2 is true.
- Statement 3 is true.
- Statement 4 is false.
The statements that are true are 1, 2, and 3.
Important Questions from Metric Spaces
Let (X, d) be a compact metric space. Let T ∶ X → X be a continuous function satisfying infn∈ℕ d(Tn(x),Tn (y)) ≠ 0 for every x, y ∈ X with x ≠ y. Then which of the following statements are true?
For any two metric spaces (X, dX), (Y, dY) a map f ∶ X → Y is said to be a closed map if whenever F is closed in X, then f(F) is closed in Y. For any subset B of a metric space, B is given the induced metric. The metric on X × Y is given by d((x, y), (x', y')) = max{dX(x, x'), dY(y, y')}. Which of the following are true?
Let X = \(\rm\displaystyle\prod_{n=1}^{\infty}\)[0, 1], that is, the space of sequences {xn}n ≥ 1 with xn ∈ [0, 1], n ≥ 1.
Define the metric d: X × X → [0, ∞) by
d({xn}n≥1, {yn}n≥1) = \(\rm\displaystyle\sup _{n ≥ 1} \frac{\left|x_n-y_n\right|}{2^n}\).
Which of the following statements are true?
If f is continuous mapping of compact metric space X into a metric space Y then:
Which statement states that "Every complete metric space is of second category"?
