Topology

2011 midterm exam in Topology

1. Suppose \(A \subseteq B\) and B is finite, show that A is also finite.

Solution: B is finite implies that there is a natural number \(n\) such that there exists a one-to-one correspondence between the points of B and the numbers {1, 2, 3, …, n}. This further implies that \(|B| = n\). Since \(A \subseteq B\), then \(|A| \leq |B|\), implying further that there is a natural number \(m = |A| \leq n\). Hence, the elements of A can be made to be mapped (one-to-one) to the elements of the set {1, 2, 3, …, m}. Therefore, A is also finite.

2. Show that the interval \((a, b]\) is neither open nor closed in \(\mathbb{R}\).

Solution: Let \(H = (a, b] = \{x \in \mathbb{R}:a \le x \leq b\}\) and \(x \in B(b, \epsilon).\) This means that \(x \in [(b-\epsilon, b+\epsilon)=(b-\epsilon)\bigcup(b, b+\epsilon)]\). Note that in the case when \(x \in (b, b+\epsilon), implies that x \notin H\) since b is the supremum of H. Hence \(B(b, \epsilon) \nsubseteq H\). Therefore H is not open. To show that is not closed, we will show that \(\mathbb{R}\setminus H\) is not open. \(\mathbb{R}\setminus H = (-\infty, a] \bigcup (b, \infty)\). Let \(x \in B(a, \epsilon)\). This means that \(x \in [(a-\epsilon, a + \epsilon) = (a-\epsilon, a] \bigcup (a, a+\epsilon)]\). Note that in the case when \(x \in (a, a+\epsilon)\), implies that \(x \notin \mathbb{R}\setminus H\) since a is the supremum of the set \((-\infty, a]\). Hence \(B(a, \epsilon) \nsubseteq \mathbb{R}\setminus H\). Therefore \(\mathbb{R}\setminus H\) is not open. Therefore the interval \((a, b]\) is neither open nor closed in \(\mathbb{R}\).

3. Suppose X, Y, and Z are metric spaces and \(f: X \to Y\) and \(g: Y \to Z\) are continuous functions Prove that \(g \circ f : X \to Z\) is also continuous.

Solution: Let \(f: X \to Y\) be continuous at x and \(g: Y \to Z\) be continuous at f(x). We will show that \(g \circ f : X \to Z\) is continuous at x. Let \(x \in X\) and let V be a neighborhood of \((g \circ f) (x)\). We need to find a neighborhood U of x such that \((g \circ f) (U) \subset V\). By the continuity of g at f(x), there exists a neighborhood W of f(x) such that \(g(W) \subset V\). By the continuity of f at x, there exists a neighborhood U of x such that \(f(U) \subset W\). Thus, \(g(f(U)) \subset g(W) \subset V\). Hence, \(g \circ f: X \to Z\) is continuous at x. Therefore \(g \circ f: X \to Z\) is also continuous.

4. Let \(f: X \to Y\) be a function from a nonempty set X into a topological space (Y, U). Further, let \(\tau\) be a class of inverses of open subsets of \(Y: \tau = \{f^{-1}[G]:G \in U\}\). Show that \(\tau\) is a topology on X.

Solution: Since U is a topology on Y, \(\varnothing \in U\). Write \(\varnothing = f^{-1}[\varnothing] \in \tau\). Hence, \(\varnothing \in \tau\).

Since U is a topology on Y, \(Y \in U\). With f a function, we can write \(X = f^{-1}[Y] \in \tau\). Hence, \(X \in \tau\).

Let \(G_1, G_2 \in U\). Since U is a topology on Y, \(G_1 \cap G_2 \in U\). Write \(f^{-1}[G_1] \cap f^{-1}[G_2] \subseteq f^{-1}[G_1 \cap G_2] \in \tau\).

Let \(G_1, G_2, G_3, \ldots \in U\). Since U is a topology on Y, \(\bigcup G_i \in U\). Write \(f^{-1}[G_1] \cup f^{-1}[G_2] \cup f^{-1}[G_3] \cap \ldots \subseteq f^{-1}[G_1 \cup G_2 \cup G_3 \cup \ldots] \in \tau\).

Therefore, \(\tau\) is a topology on X.

5. Prove that if \(X = \{0, 1\}\) and \(\tau = \{\varnothing, X, \{0\} \}\), then \(\tau\) is a topology on X.

Solution:

i. \(\varnothing, X, \in \tau\)

ii. \(\varnothing \cap X = \varnothing \in \tau\), \(\varnothing \cap \{0\} = \varnothing \in \tau\), and \(X \cap \{0\} = \{0\} \in \tau\)

iii. \(\varnothing \cup X = X \in \tau\), \(\varnothing \cup \{0\} = \{0\} \in \tau\), and \(X \cup \{0\} = X \in \tau\)

\(\therefore\) \(\tau\) is a topology on X.

6. Suppose X is a set. The discrete metric on X is defined by

\[d(x, y) = \begin{cases} 1, & \text{if } x \neq y \\ 0, & \text{if } x = y \end{cases}\]

Verify that d is a metric.

Solution:

a. As defined above, d(x, y) = 0 \(\iff\) x = y.

b. Case 1: When x = y, d(x, y) = 0 = d(y, x).

Case 2: When \(x \neq y\), d(x, y) = 1 = d(y, x).

c. Case 1: When x = y and y = z, d(x, y) = 0 \(\leq\) 0 + 0 = d(x, z) + d(z, y). Hence, \(d(x, y) \leq d(x, z) + d(z, y)\)

Case 2: When \(x \neq y\) and \(y \neq z\), d(x, y) = 1 \(\leq 2\) = 1 + 1 = d(x, z) + d(z, y). Hence, \(d(x, y) \leq d(x, z) d(z, y)\).

\(\therefore\) d is a metric on X.