1. Metric Spaces: Basic Definitions
A metric space is a mathematical structure that generalizes the concept of distance between two points. Formally, a metric space is an ordered pair \((X, d)\), where \(X\) is a non-empty set and \(d: X \times X \rightarrow \mathbb{R}\) is a function (called the metric or distance function) satisfying the following axioms for all \(x, y, z \in X\):
1. Positivity: \(d(x, y) \ge 0\), and \(d(x, y) = 0 \iff x = y\) (identity of indiscernibles).
2. Symmetry: \(d(x, y) = d(y, x)\).
3. Triangle Inequality: \(d(x, y) \le d(x, z) + d(z, y)\).
Common examples include:
• Standard Euclidean Metric on \(\mathbb{R}^n\): \(d(x, y) = \sqrt{\sum (x_i - y_i)^2}\).
• Discrete Metric: \(d(x, y) = 0\) if \(x = y\), and \(d(x, y) = 1\) if \(x \ne y\) (applicable to any set).
• Taxicab Metric (\(L_1\)): \(d(x, y) = \sum |x_i - y_i|\).
2. Compactness: Open Covers & Sequences
Compactness is a topological property that generalizes the concept of closed and bounded intervals in \(\mathbb{R}\). In higher mathematics, we define it using open sets:
Definition (Compact): \(K\) is compact if every open cover of \(K\) contains a finite subcover.
In metric spaces, compactness is exceptionally well-behaved and is completely equivalent to sequential compactness:
A subset \(K\) of a metric space is compact if and only if every sequence \(\{x_n\}\) in \(K\) contains a subsequence \(\{x_{n_k}\}\) that converges to a point in \(K\).
3. The Heine-Borel Theorem & Boundary Conditions
One of the most famous results in real analysis is the Heine-Borel Theorem:
A subset \(K \subset \mathbb{R}^n\) (with the standard Euclidean metric) is compact if and only if it is closed and bounded.
WARNING: This equivalence is specific to \(\mathbb{R}^n\). In general metric spaces, a closed and bounded set is not necessarily compact. For example, in an infinite discrete metric space, the entire space is closed and bounded (maximum distance is 1) but is not compact because the open cover of singletons has no finite subcover (see Solved Question 1 below).
4. Solved CSIR-NET & GATE Questions
Question 1 (CSIR-NET Mathematics)
Let \(X = \mathbb{R}\) be equipped with the discrete metric: \[d(x, y) = \begin{cases} 0 & \text{if } x = y \\ 1 & \text{if } x \neq y \end{cases}\] Determine whether the closed interval \(K = [0, 1]\) is compact under this metric.
Detailed Solution:
- Understand the nature of open sets in a discrete metric space:
An open ball of radius \(r=1/2\) around any point \(x\) is: \[B(x, 1/2) = \{y \in X \mid d(x,y) < 1/2\}\] Since the only distance values are 0 and 1, the only point at distance less than 1/2 is \(x\) itself. \[B(x, 1/2) = \{x\}\] Since open balls are open sets, every singleton set \(\{x\}\) is an open set in \((X, d)\). - Construct an open cover for \(K = [0, 1]\):
Since singletons are open, we can cover \(K\) by the union of all its singletons: \[K \subset \bigcup_{x \in [0, 1]} \{x\}\] This collection of open sets \(\{ \{x\} \}_{x \in [0, 1]}\) forms an open cover of \(K\). - Analyze the subcover requirement:
Since \([0, 1]\) contains infinitely many points, any finite subcollection of singletons \(\{ \{x_1\}, \{x_2\}, \dots, \{x_m\} \}\) can only cover \(m\) points.
Thus, it cannot cover the entire interval \([0, 1]\). - Conclusion:
We have constructed an open cover of \(K\) that contains no finite subcover.
Therefore, the closed and bounded interval \([0, 1]\) is not compact under the discrete metric.
Answer: No, \(K = [0, 1]\) is not compact under the discrete metric.
Question 2 (GATE Mathematics)
Let \((X, d)\) and \((Y, \rho)\) be metric spaces, and let \(f: X \rightarrow Y\) be a continuous function. Prove that if \(K \subset X\) is compact, then its image \(f(K) \subset Y\) is also compact.
Detailed Solution:
- Let \(\{V_\alpha\}\) be an arbitrary open cover of \(f(K)\) in \(Y\): \[f(K) \subset \bigcup_{\alpha} V_\alpha\]
- Use the continuity property of \(f\):
Since \(f\) is continuous, the preimage of any open set in \(Y\) is an open set in \(X\).
Therefore, \(U_\alpha = f^{-1}(V_\alpha)\) is open in \(X\) for every \(\alpha\). - Show that \(\{U_\alpha\}\) covers \(K\):
For any \(x \in K\), we have \(f(x) \in f(K)\).
Since \(\{V_\alpha\}\) covers \(f(K)\), there exists some index \(\beta\) such that \(f(x) \in V_\beta\).
By definition of preimage, this means \(x \in f^{-1}(V_\beta) = U_\beta\).
Thus, \(K \subset \bigcup_{\alpha} U_\alpha\), meaning \(\{U_\alpha\}\) is an open cover of \(K\). - Apply the compactness of \(K\):
Since \(K\) is compact, the open cover \(\{U_\alpha\}\) has a finite subcover, say \(\{U_{\alpha_1}, U_{\alpha_2}, \dots, U_{\alpha_n}\}\): \[K \subset \bigcup_{i=1}^n U_{\alpha_i}\] - Take the image of both sides to return to \(Y\): \[f(K) \subset f\left(\bigcup_{i=1}^n f^{-1}(V_{\alpha_i})\right) = \bigcup_{i=1}^n f(f^{-1}(V_{\alpha_i})) \subset \bigcup_{i=1}^n V_{\alpha_i}\] This finite subcollection \(\{V_{\alpha_1}, V_{\alpha_2}, \dots, V_{\alpha_n}\}\) covers \(f(K)\).
- Conclusion:
Every open cover of \(f(K)\) contains a finite subcover. Thus, \(f(K)\) is compact.
Answer: The theorem is proved: the continuous image of a compact set is compact.
Adarsh V
Senior Mathematics Faculty | IIT Bombay Alumni | GATE AIR-20Adarsh V holds an MSc in Mathematics from IIT Bombay and a BSc from Chennai Mathematical Institute. He secured GATE AIR 20 and specializes in Real Analysis, Topology, and Metric Space theory at Benzil Academy.