Exercises:
- Let \( X = \{a,b,c,d,e,f \}\). Determine whether (or) not each of the following collection of subsets of \(X\) is a topology on \(X\)
- \( \tau_1 = \{X, \emptyset, \{a\}, \{a,f\}, \{b,f\}, \{a,b,f\} \} \)
- Note that \( \{a,f\} \bigcap \{b,f\} = \{f\} \notin \tau_1 \). Hence, \( \tau_1\) is not a topology.
- \( \tau_2 = \{X, \emptyset, \{a,b,f\}, \{a,b,d\}, \{a,b,d,f\} \} \)
- Again, note that \(\{a,b,f\}\bigcap \{a,b,d\} =\{a,b\} \notin \tau_2\). Hence, \( \tau_2\) is not a topology.
- \( \tau_3 = \{X, \emptyset, \{f\}, \{e,f\}, \{a,f\} \}\)
- Note that \(\{e,f \} \bigcup \{a,f\} = \{a,e,f\} \notin \tau_3\}\). Hence, \( \tau_3\) is not a topology.
- Let \(X = \{a,b,c,d,e,f\}\). Which of the following collections of subsets of \(X\) is a topology on \(X\)? (Justify your answers)
- \(\tau_1 = \{X, \emptyset, \{c\}, \{b,d,e\}, \{b,c,d,e\}, \{b\} \}\)
- Not a topology since \( \{c\} \bigcup \{b\} = \{b,c\} \notin \tau_1\)
- \(\tau_2 = \{X, \emptyset, \{a\}, \{b,d,e\}, \{a,b,d\}, \{a,b,d,e\} \}\)
- Not a topology since \(\{b,d,e\}\bigcup \{a,b,d\} = \{b,d\} \notin \tau_2\)
- \(\tau_3 = \{X, \emptyset, \{b\}, \{a,b,c\}, \{d,e,f\}, \{b,d,e,f\} \}\)
- It is a topology since
- It contains \(X\) and \(\emptyset\).
- Any union of sets in \(\tau_3\) is in \(\tau_3\)
- Any intersection of sets in \(\tau_3\) is in \(\tau_3\)
- Let \(X = \{a,b,c,d,e,f \}\) and \(\tau\) is the discrete topology on \(X\). Which of the following statements are true?
- \(X \in \tau\) True
- \(\{X\} \in \tau\) False
- \(\{\emptyset\} \in \tau\) False
- \(\emptyset \in \tau\) True
- \(\emptyset \in X\) False
- \(\{\emptyset\} \in X\) False
- \(\{a\} \in \tau\) True
- \(a \in \tau\) False
- \(\emptyset \subseteq X\) True
- \(\{a\} \in X\) False
- \(\{\emptyset\} \subseteq X\) False
- \(a \in X\) True
- \(X \subseteq \tau\) False
- \(\{a\} \subseteq \tau\) False
- \(\{X\} \subseteq \tau\) True
- \(a \subseteq \tau\) False
- Let \((X,\tau)\) be any topological space. Verify that the intersection of any finite number of members of \(\tau\) is a member of \(\tau\).
- Let \(P_n\) be the statement that "If \(\{A_i\}_{i=1}^n \in \tau \implies \bigcap_{i=1}^n A_i \in \tau \)". Let \(S = \{n \in \mathbb{N}: P_n \text{ is a true statement}\}\). From the definition of a topology, we have \(1 \in \mathbb{N}\). Assume that \(k \in S\) i.e. \(P_k\) is a true statement i.e. whenever \(\{A_i\}_{i=1}^k \in \tau \implies \bigcap_{i=1}^k A_i \in \tau \). Consider \(\{A_i\}_{i=1}^{k+1} \in \tau \). \( \bigcap_{i=1}^{k+1} A_i = \left( \bigcap_{i=1}^{k} A_i \right) \bigcap A_{k+1}\). By assumption, we have \(B = \left( \bigcap_{i=1}^{k} A_i \right) \in \tau\). Now from the definition of a topology, we get \(B \bigcap A_{k+1} \in \tau\) since \(B,A_{k+1} \in \tau\). Hence, we get \( \bigcap_{i=1}^{k+1} A_i \in \tau \) whenever \(\{A_i\}_{i=1}^{k+1} \in \tau \). Hence, \(P_{k+1}\) is a true statement. Hence, \( k+1 \in S\). So by principle of mathematical induction we have \( 1 \in S \), and \(k+1 \in S\) whenever \( k \in S \). Hence, \( S = \mathbb{N} \). Hence, the intersection of any finite number of members of \(\tau\) is a member of \(\tau\).
- Let \(\mathbb{R}\) be the set of all real numbers. Prove that each of the following collections of subsets of \(\mathbb{R}\) is a topology
- \(\tau_1\) consists of \(\mathbb{R}, \emptyset\) and every interval \((-n,n)\), for any positive integer
- Clearly, \(\mathbb{R}, \emptyset \in \tau_1\)
- First note that \(\tau_1\) is a countable set and hence all we are interested in is in finite (or) countably infinite unions. Let \(A_{k} = (-k,k)\). Note that \(A_k\)'s are monotone increasing sequence of sets. Hence, any finite union of the form \(\bigcup_{l=1}^{m} A_{k_l} = A_{p}\) where \(p=\max(k_1,k_2,\ldots,k_m)\). And \(A_{p} \in \tau_1\) and hence any finite union again belongs to \(\tau_1\). Any infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_l \in \mathbb{N}\) can be rewritten either as a finite union of the form \(\bigcup_{l=1}^{m} A_{k_l}\) (or) an infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_i \neq k_j\). The claim now is that any infinite union where \(k_i \neq k_j\) is \(\mathbb{R}\) i.e. \(\bigcup_{l=1}^{\infty} A_{k_l} = \mathbb{R}\) where \(k_i \neq k_j\). To prove this, we prove the two way inclusion. Note that \(A_{k_l} \in \mathbb{R}\). Hence, \(\bigcup_{l=1}^{\infty} A_{k_l} \subseteq \mathbb{R}\). Further given any \(x \in \mathbb{R}\), by archimedian property \( \exists n \in \mathbb{N}\) such that \(x \in A_n\). Further since it is an infinite union, \( \exists k_l\) such that \( x \in A_n \subseteq A_{k_l}\). Hence,
- Now we need to prove the last claim that intersection of any two elements of \(\tau_1\) gives an element in \(\tau_1\). Consider \(A_m,A_n \in \tau_1\). By well-ordering principle, we get \(m \lt n\)
- \(\tau_2\) consists of \(\mathbb{R}, \emptyset\) and every interval \([-n,n]\), for any positive integer
- Clearly, \(\mathbb{R}, \emptyset \in \tau_2\)
- First note that \(\tau_2\) is a countable set and hence all we are interested in is in finite (or) countably infinite unions. Let \(A_{k} = [-k,k]\). Note that \(A_k\)'s are monotone increasing sequence of sets. Hence, any finite union of the form \(\bigcup_{l=1}^{m} A_{k_l} = A_{p}\) where \(p=\max(k_1,k_2,\ldots,k_m)\). And \(A_{p} \in \tau_2\) and hence any finite union again belongs to \(\tau_2\). Any infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_l \in \mathbb{N}\) can be rewritten either as a finite union of the form \(\bigcup_{l=1}^{m} A_{k_l}\) (or) an infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_i \neq k_j\). The claim now is that any infinite union where \(k_i \neq k_j\) is \(\mathbb{R}\) i.e. \(\bigcup_{l=1}^{\infty} A_{k_l} = \mathbb{R}\) where \(k_i \neq k_j\). To prove this, we prove the two way inclusion. Note that \(A_{k_l} \in \mathbb{R}\). Hence, \(\bigcup_{l=1}^{\infty} A_{k_l} \subseteq \mathbb{R}\). Further given any \(x \in \mathbb{R}\), by archimedian property \( \exists n \in \mathbb{N}\) such that \(x \in A_n\). Further since it is an infinite union, \( \exists k_l\) such that \( x \in A_n \subseteq A_{k_l}\). Hence,
- Now we need to prove the last claim that intersection of any two elements of \(\tau_2\) gives an element in \(\tau_2\). Consider \(A_m,A_n \in \tau_2\). By well-ordering principle, we get \(m \lt n\)
- \(\tau_3\) consists of \(\mathbb{R}, \emptyset\) and every interval \([n,\infty)\), for any positive integer
- Clearly, \(\mathbb{R}, \emptyset \in \tau_3\)
- First note that \(\tau_3\) is a countable set and hence all we are interested in is in finite (or) countably infinite unions. Let \(A_{k} = [k,\infty)\). Note that \(A_k\)'s are monotone decreasing sequence of sets. Hence, any finite union of the form \(\bigcup_{l=1}^{m} A_{k_l} = A_{p}\) where \(p=\min(k_1,k_2,\ldots,k_m)\). And \(A_{p} \in \tau_3\) and hence any finite union again belongs to \(\tau_3\). Any infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_l \in \mathbb{N}\) equals \(A_p\) where \(p = \min(k_1,k_2,\ldots) \).
- Now we need to prove the last claim that intersection of any two elements of \(\tau_3\) gives an element in \(\tau_3\). Consider \(A_m,A_n \in \tau_3\). By well-ordering principle, we get \(m \lt n\)
- Let \(\mathbb{N}\) be the set of all positive integers. Prove that each of the following collections of subsets of \(\mathbb{N}\) is a topology.
- \(\tau_1\) consists of \(\mathbb{N}, \emptyset\) and every set of the form \(\{1,2,\ldots,n\}\) for any positive integer \(n\). This is called the initial segment topology.
- Argument similar to the above question
- \(\tau_2\) consists of \(\mathbb{N}, \emptyset\) and every set of the form \(\{n,n+1,\ldots\}\) for any positive integer \(n\). This is called the final segment topology.
- Argument similar to the above question
- List all possible topologies on the following sets:
- \(X = \{a,b\}\)
- \(\tau_1 = \{\emptyset,X\} \)
- \(\tau_2 = \{\emptyset,\{a\},X\} \)
- \(\tau_3 = \{\emptyset,\{b\},X\} \)
- \(\tau_4 = \{\emptyset,\{a\},\{b\},X\} \)
- \(Y = \{a,b,c\}\)
- \(\tau_1 = \{\emptyset,X\} \)
- \(\tau_2 = \{\emptyset,\{a\},X\} \)
- \(\tau_3 = \{\emptyset,\{b\},X\} \)
- \(\tau_4 = \{\emptyset,\{c\},X\} \)
- \(\tau_5 = \{\emptyset,\{a\},\{b\},\{a,b\},X\} \)
- \(\tau_6 = \{\emptyset,\{c\},\{b\},\{c,b\},X\} \)
- \(\tau_7 = \{\emptyset,\{a\},\{c\},\{a,c\},X\} \)
- \(\tau_8 = \{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{b,c\},\{c,a\},X\} \)
- Let \(X\) be an infinite set and \(\tau\) be a topology on \(X\). If every infinite subset of \(X\) is in \(\tau\), prove that \(\tau\) is the discrete topology.
- The idea is to prove that every singleton set is in the topology. By a proposition proved earlier, it follows that \(\tau\) is the discrete topology. To prove this, first observe that given an infinite set, \(X\), we can choose an infinite set \(A\) such that \(X \backslash A\) is also infinite.
- For instance, if the given set \(X\) is countably infinite, then we can set a bijection from \(\mathbb{N}\) to the set \(X\) i.e. we can write the set \(X\) as \(\{x_k: k \in \mathbb{N}\}\). Let \(A = \{x_k: k \in \mathbb{Z}^+_{odd}\}\). Clearly, both \(A\) and \(X \backslash A\) are countably infinite subsets of \(X\).
- If the given set \(X\) is uncountably infinite, then by axiom of choice, it is possible to choose a countably infinite subset, say \(A\) of \(X\). Hence, \(X \backslash A\) is also an infinite set, infact it is uncountably infinite. (I was initially unsure if this could be done. But thanks to math.stackexchange, Arturo Magidin told me that this was possible with the "Axiom of choice".)
- Now we have \(A\) and \(X \backslash A\) to be infinite sets. For any \(x \in X\), consider \(A \cup \{x\}\) and \(X \backslash A \cup \{x\}\). These two are infinite sets and hence both the sets are in the topology \(\tau\) and hence the intersection of the two sets must also be in the topology \(\tau\). The intersection of the above two sets is nothing but \(\{x\}\). This is true for every set \(\{x\}\). Hence, every singleton set belongs to the topology and hence the topology is discrete.
- Let \(\mathbb{R}\) be the set of all real numbers. Precisely three of the following collections of subsets of \(\mathbb{R}\) are topologies? Identify these and justify your answer
- \(\tau\) consists of \(\mathbb{R}, \emptyset\) and every interval \((a,b)\), for \(a\) and \(b\) any real numbers with \(a \lt b\)
- Not a topology since for instance \((1,2), (3,4) \in \tau\) but \((1,2) \bigcup (3,4) \notin \tau\)
- \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \((-r,r)\) for \(r \in \mathbb{R}\)
- Is a topology by argument similar to question \(6\)
- \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \((-r,r)\) for \(r \in \mathbb{Q}^+\)
- Not a topology since if we take rationals \(\{x_n\}_{n=1}^{\infty} \) such that \(x_{n+1} = \frac{2x_n}{2x_n^2+1} \) with \(x_1 = \frac12\) and let \(A_n = (-x_n,x_n)\) and consider \( \bigcup_{k=1}^{\infty}A_k \). Note that \(\{x_k\}_{k=1}^{\infty}\) is an increasing sequence bounded above and converges to \(\frac1{\sqrt{2}}\). This means that the sets \(A_k\) are monotone increasing sequence of sets, and \(A_k \in \tau\). \( \bigcup_{k=1}^{\infty}A_k = \left(-\frac1{\sqrt{2}},\frac1{\sqrt{2}} \right) \notin \tau \)
- \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \([-r,r]\) for \(r \in \mathbb{Q}^+\)
- Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = 1-\frac1n\). Let \(A_n = [-x_n,x_n]\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-1,1 \right) \notin \tau \)
- \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \((-r,r)\) for \(r \in \mathbb{R}^+ \backslash \mathbb{Q} \)
- Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = 1- \frac{\sqrt{2}}{2n}\). Clearly, \(x_n \in \mathbb{R}^+ \backslash \mathbb{Q}\), \(\forall n \in \mathbb{N}\). Let \(A_n = (-x_n,x_n)\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-1,1 \right) \notin \tau \)
- \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \([-r,r]\) for \(r \in \mathbb{R}^+ \backslash \mathbb{Q} \)
- Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = \sqrt{2}- \frac{1}{n}\). Clearly, \(x_n \in \mathbb{R}^+ \backslash \mathbb{Q}\), \(\forall n \in \mathbb{N}\). Let \(A_n = [-x_n,x_n]\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-\sqrt{2},\sqrt{2} \right) \notin \tau \)
- \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \([-r,r)\) for \(r \in \mathbb{R}^+ \)
- Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = 1- \frac1{n}\). Clearly, \(x_n \in \mathbb{R}^+\), \(\forall n \in \mathbb{N}\). Let \(A_n = [-x_n,x_n)\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-1,1 \right) \notin \tau \)
- \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \((-r,r]\) for \(r \in \mathbb{R}^+ \)
- Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = 1- \frac1{n}\). Clearly, \(x_n \in \mathbb{R}^+\), \(\forall n \in \mathbb{N}\). Let \(A_n = (-x_n,x_n]\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-1,1 \right) \notin \tau \)
- \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \([-r,r]\) and every interval \((-r,r)\) for \(r \in \mathbb{R}^+ \)
- Is a topology
- Clearly, \(X,\emptyset \in \tau\)
- Let \(A^{(0)}_{r}=(-r,r)\) and \(A^{(1)}_r = [-r,r]\). Consider \(A = \left (\bigcup_{x \in \Gamma_0} A^{(0)}_x \right) \bigcup \left( \bigcup_{y \in \Gamma_1} A^{(1)}_y \right) \). Then \(A = A^{(0)}_x\) for some \(x \in \mathbb{R}^+\) (or) \(A = A^{(1)}_y\) for some \(y \in \mathbb{R}^+\) (or) \(A=\mathbb{R}\).
- Intersection of any two sets is closed in \(\tau\)
- \(\tau\) consists of \(\mathbb{R}, \emptyset\), every interval \([-n,n]\), and every interval \((-r,r)\), for \(n \in \mathbb{Z}^+\) and \(r \in \mathbb{R}^+\)
- Is a topology. Again check the definitions.
No comments:
Post a Comment