Topological Spaces: Basic Concepts
Definition Given a set \(X\), a topology in \(X\) is a system \(\tau\) of subsets \(G \subseteq X\), called open sets (relative to \(\tau\)), with the following two properties:
\(X\) and \(\emptyset\) belong to \(\tau\);
arbitrary unions \(\cup_\alpha G_\alpha\) and finite intersections \(\cap_{k=1}^n G_k\) of open sets belong to \(\tau\).
Definition A topological space is a pair \((X, \tau)\) consisting of a set \(X\) and a topology \(\tau\) defined in \(X\).
More definitions Elements of a topological space are called points. Closed sets are the complements of open sets. A neighbourhood of a point \(x\) is any open set containing the point. A point \(x \in T\) is a contact point of a set \(M \subseteq T\) if every neighbourhood of \(x\) contains at least one point of \(M\). A point \(x \in T\) is a limit point of a set \(M \subseteq T\) if every neighbourhood of \(x\) contains infinitely many points of \(M\). The set of all contact points of a set \(M \subseteq T\) is called a closure of \(M\), denoted by \([M]\).
Note Example 4, in which \(T = \{a, b\}\) and the open sets are \(\emptyset\), \(\{b\}\) and \(T\), with \([T] = \{b\}\), demonstrates that the closure is the set of all contact points and not limit points.
Definition Let \(\tau_1\) and \(\tau_2\) be two topologies defined in the same set \(X\). Then \(\tau_1\) is stronger than \(\tau_2\) (or equivalently \(\tau_2\) is weaker than \(\tau_1\)) if \(\tau_2 \subseteq \tau_1\), i.e. every set of the system \(\tau_2\) is a set of the system \(\tau_1\).
Theorem 1
The intersection \(\tau = \cap_\alpha \tau_\alpha\) of any sets of topologies in \(X\) is itself a topology in \(X\).
Proof
Clearly \(\cap_\alpha \tau_\alpha\) contains \(X\) and \(\emptyset\). Also, every \(\tau_\alpha\) is closed (algebraically) under the operations of taking arbitrary unions and finite intersections, the same is true of \(\cap_\alpha \tau_\alpha\).
Corollary
Let \(\mathscr{B}\) be any system of subsets of a set \(X\). Then there exists a minimal topology in \(X\) containing \(\mathscr{B}\), i.e. a topology \(\tau(\mathscr{B})\) containing \(\mathscr{B}\) and contained in every topology containing \(\mathscr{B}\).
Proof
A topology containing \(\mathscr{B}\) always exists, e.g. the topology in which every subset of \(X\) is open. The intersection fo all topologies containing \(\mathscr{B}\) is the desired minimal topology \(\tau(\mathscr{B})\), often called the topology generated by the system \(\mathscr{B}\).
More definitions Let \(\mathscr{B}\) be a system of subsets of \(X\), and \(A\) a fixed subset of \(X\). Then the trace of the system \(\mathscr{B}\) on the set \(A\) is the system \(\mathscr{B}_A\) consisting of all subsets of \(X\) of the form \(A \cap B\), \(B \in \mathscr{B}\). The trace (on A) of a topology \(\tau\) (defined in \(X\)) is a topology \(\tau_A\) in \(A\), often called a relative topology. In this sense, every subset \(A\) of a given topological space \((X, \tau)\) generates a new topological space \((A, \tau_A)\) called a subspace of the original topological space \((X, \tau)\).
Definition A family \(\mathscr{G}\) of open subsets of a topological space \(T\) is called a base for \(T\) if every open set in \(T\) can be represented as a union of sets in \(\mathscr{G}\).
Theorem 2
Given a set \(T\), let \(\mathscr{G}\) be a system of subsets \(G_\alpha \subseteq T\) with the following two properties:
every point \(x \in T\) belongs to at least one \(G_\alpha \in \mathscr{G}\);
if \(x \in G_\alpha \cap G_\beta\), then there is a \(G_\gamma \in \mathscr{G}\) such that \(x \in G_\gamma \in G_\alpha \cap G_\beta\).
Suppose the empty set \(\emptyset\) and all sets representable as unions of sets \(G_\alpha\) are designated as open. Then \(T\) is a topological space, and \(\mathscr{G}\) is a base for \(T\).
Proof
Evident from the conditions of the theorem that the whole set \(T\) and the empty set \(\emptyset\) are open sets; and the union of any number of open sets is open. Just have to show that the intersection of a finite number of open sets is open. It is enough to prove this for just two (arbitary) sets. Let \(A = \cup_\alpha G_\alpha\), \(B = \cup_\beta G_\beta\). Then \(A \cap B = \cup_{\alpha, \beta} (G_\alpha \cap G_\beta)\) (1). By hypothesis, for any point \(x \in G_\alpha \cap G_\beta\) there exists a \(G_\gamma \in \mathscr{G}\) such that \(x \in G_\gamma \subseteq G_\alpha \cap G_\beta\). Hence the set \(G_\alpha \cap G_\beta\) is open, being the union of all \(G_\gamma\) contained in \(G_\alpha \cap G_\beta\). But then (1) is also open. Thus \(T\) is a topological space. The fact that \(\mathscr{G}\) is a base for \(T\) is clear from the way open sets in \(T\) are defined.
Theorem 3
A system \(\mathscr{G}\) of open sets \(G_\alpha\) in a topological space \(T\) is a base for \(T\) if and only if, given any open set \(G \subseteq T\) and any point \(x \in G\), there is a set \(G_\alpha \in \mathscr{G}\) such that \(x \in G_\alpha \subseteq G\).
Proof
If \(\mathscr{G}\) is a base for \(T\), then every open set \(G \subseteq T\) is set \(G \subseteq T\) and any point \(x \in G\), there is a set \(G_\alpha \in \mathscr{G}\) such that \(x \in G_\alpha \subseteq G\). union \(G = \cup_\alpha G_\alpha\) of sets \(G_\alpha \in \mathscr{G}\). Therefore every point \(x \in G\) is contained in some set \(G_\alpha \subseteq G\). Conversely given anyten set \(G \subseteq T\), suppose that for every point \(x \in G\) there is a set \(G_\alpha(x) \in \mathscr{G}\) such that \(x \in G_\alpha(x) \subseteq G\). Then \(G = \cup_{x \in G} G_\alpha (x)\), i.e. \(G\) is a union of sets in \(\mathscr{G}\).
More definitions Topological spaces with a countable base have at least one base containing no more than countably many sets. Such a space is said to satisfy the second axiom of countability.
Theorem 4
If a topological space \(T\) has a countable base, then \(T\) contains a countable everywhere dense subset, i.e. a countable set \(M \subseteq T\) such that \([M] \subseteq T\).
Proof
Let \(\mathscr{G} = \{G_1, \ldots, G_n, \ldots\}\) be a countable base for \(T\), and choose a point \(x_n\) in each \(G_n\). Then the set \(M = \{x_1, x_2, \ldots, x_n, \ldots\}\) is countable. Moreover, \(M\) is everywhere dense in \(T\), since otherwise the nonempty open set \(G = T - [M]\) would contain no points of \(M\). But this is impossible, since \(G\) is a union of some of the sets \(G_n\) in \(\mathscr{G}\) and \(G_n\) contains the point \(x_n \in M\).
Theorem 5
If a metric space \(R\) has a countable everywhere dense subset, then \(R\) has a countable base.
Proof
Suppose \(R\) has a countable everywhere dense subset \(\{x_1, \ldots, x_n, \ldots\}\). Then, given any open set \(G \subseteq R\) and any \(x \in G\), there is an open sphere \(S(x_m, \frac{1}{n})\) such that \(x \in S(x_m, \frac{1}{n}) \subseteq G\) for suitable positive integers \(m\) and \(n\) (why?). Hence the open sphere \(S(x_m, \frac{1}{n})\), where \(m\) and \(n\) range over all positive integers, form a countable base for \(R\).
Note Combining Theorems 4 and 5, we see that a metric space \(R\) has a countable base if and only if it has a countable everywhere dense set.
More definitions A system \(\mathscr{M}\) of sets \(M_\alpha\) is a cover (or covering) of a topological space \(T\), and \(\mathscr{M}\) is said to cover \(T\), if \(T = \cup_\alpha M_\alpha\). A cover consisting of open (or closed) sets only is called an open (or closed) cover. If \(\mathscr{M}\) is a cover of a topological space \(T\), then by a subcover of \(\mathscr{M}\) we mean any subset of \(\mathscr{M}\) which also covers \(T\).
Theorem 6
If \(T\) is a topological space with a countable base \(\mathscr{G}\), then every open cover \(\mathscr{O}\) has a finite or countable subcover.
Proof
Since \(\mathscr{O}\) covers \(T\), each point \(x \in T\) belongs to some open set \(O_\alpha \in \mathscr{O}\). Moreover, since \(\mathscr{G}\) is a countable base for \(T\), for each \(x \in T\) there is a set \(G_n(x)\) such that \(x \in G_n (x) \subseteq O_\alpha\). The collection of all sets \(G_n (x)\) selected in this way is finite or countable and covers \(T\). For each \(G_n (x)\) we now choose one of the sets \(O_\alpha\) containing \(G_n(x)\), thereby obtaining a finite or countable subcover of \(\mathscr{O}\).
More definitions A topological space \(T\) is connected if it has no subsets other than \(\emptyset\) and \(T\) which are both open and closed.
Theorem 7
If a topological space \(T\) satisfies the first axiom of countability, i.e. if there is a countable neighbourhood base at every point \(x \in T\), then every contact point \(x\) of a set \(M \subseteq T\) is the limit of a convergent sequence of points in \(M\).
Proof
Let \(\mathscr{O}\) be a countable neighbourhood base at \(x\), consisting of sets \(O_n\). it can be assumed that \(O_{n+1} \subseteq O_n (n = 1, 2, \ldots)\), since otherwise we need only replace \(O_n\) by \(\cap_{k=1}^n O_k\). Let \(x_n\) be any point of \(M\) contained in \(O_n\). Such a point \(x_n\) can always be found, since \(x\) is a contact point of \(M\). Then the sequence \(\{x_n\}\) obviously converges to \(x\).
Defintion Suppose that for each pair of distinct points \(x\) and \(y\) in a topological space \(T\), there is a neighbourhood \(O_x\) of \(x\) and a neighbourhood \(O_y\) of \(y\) such that \(x \notin O_y\), \(y \notin O_x\). Then \(T\) is said to satisfy the first axiom of separation and is called a \(T_1\)-space.
Theorem 8
Every finite subset of a \(T_1\)-space is closed.
Proof
Given any single-element set \(\{x\}\), suppose \(y \neq x\). Then \(y\) has a neighbourhood \(O_y\) which does not contain \(x\), i.e. \(y \notin [\{x\}]\). Therefore \([\{x\}] = \{x\}\), i.e. every ‘singleton’ \(\{x\}\) is closed. But every finite union of closed sets is itself closed. Hence every finite subset of the given space is closed.
Definition Suppose that for each pair of distinct points \(x\) and \(y\) in a topological space \(T\), there is a neighbourhood \(O_x\) of \(x\) and a neighbourhood \(O_y\) of \(y\) such that \(O_x \cap O_y = \emptyset\). Then \(T\) is said to satisfy the second (or Hausdorff) axiom of separation and is called a \(T_2\)-space or Hausdorff space.
Definition A \(T_1\)-space \(T\) is said to be normal if for each pair of disjoint closed sets \(F_1\) and \(F_2\) in \(T\), there is an open set \(O_1\) containing \(F_1\) and an open set \(O_2\) containing \(F_2\) such that \(O_1 \cap O_2 = \emptyset\).
Theorem 9
Every metric space is normal.
Proof
Let \(X\) and \(Y\) be any two disjoint closed subsets of \(R\). Every point \(x \in X\) has a neighbourhood \(O_x\) disjoint from \(Y\), and hence is at a positive distance \(\rho_x\) from \(Y\) (recall Problem 9 from page 54). Similarly, every point \(y \in Y\) is at a positive distance \(\rho_y\) form \(X\). Consider the open sets \(U = \cup_{x \in X} S(x, \frac{1}{2} \rho_x)\), \(V = \cup_{y \in Y} S(y, \frac{1}{2} \rho_y)\), where, as usual, \(S(x, r)\) is the open sphere with center \(x\) and radius \(r\). It is clear that \(X \subseteq U\), \(Y \subseteq V\). Moreover, \(U\) and \(V\) are disjoint. In fact, suppose to the contrary that there is a point \(z \in U \cap V\). Then there are points \(x_0 \in X\), \(y_0 \in Y\) such that \(\rho(x_0, z) < \frac{1}{2} \rho_{x_0}\), \(\rho(z, y_0) < \frac{1}{2} \rho_{y_0}\). Assume, to be explicit, that \(\rho_{x_0} \leq \rho_{y_0}\). Then \(\rho(x_0, y_0) \leq \rho(x_0, z) + \rho(z, y_0) < \frac{1}{2} \rho_{x_0} + \frac{1}{2} \rho_{y_0} \leq \rho_{y_0}\) i.e. \(x_0 \in S(y_0, \rho_y)\). This contradicts the definition of \(\rho_{y_0}\), and shows that there is no point \(z \in U \cap V\).
More definitions A property of a topologica space \(T\) shared by every subspace of \(T\) is said to be hereditary.
Note Normality is not a hereditary property.
More definitions Let \(f\) be a mapping of one topological space \(X\) into another topological space \(Y\), so that \(f\) associates an element \(y = f(x) \in Y\) with each element \(x \in X\). Then \(f\) is said to be continuous at a point \(x_0 \in X\) if, given any neighbourhood \(V_{y_0}\) of the point \(y_0 = f(x_0)\), there is a neighbourhood \(U_{x_0}\) of the point \(x_0\) such that \(f(U_{x_0}) \subseteq V_{y_0}\). \(f\) is continuous on \(X\) if it is continuous at every point of \(X\). A continuous mapping of a topological space \(X\) into the real line is called a continuous real function on \(X\).
Theorem 10
A mapping \(f\) of a topological space \(X\) into a topological space \(Y\) is continuous if and only if the preimage \(\Gamma = f^{-1}(G)\) of every open set \(G \subseteq Y\) is open (in \(X\)).
Proof
Suppose \(f\) is continuous and \(G\) is an open subset of \(Y\). Choose \(x \in \Gamma = f^{-1}(G)\) and let \(y = f(x)\). Then \(G\) is a neighbourhood of \(y\). By continuity of \(f\), tyhere is a neighbourhood \(U_x\) of \(x\) such that \(f(U_x) \subseteq G\), i.e. \(U_x \subseteq \Gamma\). In other words, every point \(x \in \Gamma\) has a neighbourhood contained in \(\Gamma\), so \(\Gamma\) is open.
Conversely, suppose \(\Gamma = f^{-1}(G)\) is open whenever \(G \subseteq Y\) is open. Let \(x \in X\) and \(V_y\) be a neighbourhood of \(y = f(x)\). Then clearly \(x \in f^{-1}(V_y)\), and \(f^{-1}(V_y)\) is open, by hypothesis. Therefore \(U_x = f^{-1}(V_y)\) is a neighbourhood of \(x\) such that \(f(U_x) \subseteq V_y\). In other words, \(f\) is continuous at \(x\) and hence on \(X\), since \(x\) is an arbitrary point of \(X\).
Theorem 10’
A mapping \(f\) of a topological space \(X\) into a topological space \(Y\) is continuous if and only if the preimage \(\Gamma = f^{-1}(F)\) of every closed set \(F \subseteq Y\) is closed (in \(X\)).
Proof
Use the fact that thereforereimage ofclosed complement is the complement of the preimage.
Theorem 11
Given topological spaces \(X\), \(Y\) and \(Z\), suppose \(f\) is a continuous mapping of \(X\) into \(Y\) and \(\phi\) is a continuous mapping of \(Y\) into \(Z\). Then the mapping \(\phi f\), i.e. the mapping carrying \(x\) into \(\phi(f(x))\), is continuous.
Proof
Immediate consequence of Theorem 10.
More definitions Given two topological spaces \(X\) and \(Y\), let \(f\) be a one-to-one mapping of \(X\) onto \(Y\), and suppose \(f\) and \(f^{-1}\) are both continuous.Then \(f\) is called a homeomorphic mapping or a homeomorphism (between \(X\) and \(Y\)). Two spaces \(X\) and \(Y\) are said to be homeomorphic if there exists a homeomorphism between them.
More definitions A topological space \(T\) is said to be metrizable if its topology can be specified by means of some metric (more exactly, if it is homemorphic to some metric space).
Urysohn’s Metrization Theorem A necessary and sufficient condition for a topological space with a countable base to be metrizable is that it be normal.
Problems
Prove that \(G \subseteq T\) is open iff every \(x \in G\) has a neighbourhood contained in \(G\).
If all \(x \in G\) has a neighbourhood contained in \(G\), then for each \(x\) let \(O_x \subseteq G\) be the open neighbourhood contained in \(G\). Then \(\cup_{x \in G} O_x \subseteq G\) is open, as topologies are closed under arbitrary unions. However \(\cup_{x \in G} O_x \supseteq G\) since \(\cup_{x \in G} O_x\) contains every point of \(G\). Hence \(G = \cup_{x \in G} O_x\) is open.
Prove
\([M] = M\) iff \(M\) closed, i.e. \(T-G\) of an open set \(G \subseteq T\).
If \([M] = M\), then \(M\) contains all of its contact points. So for all \(x \in T-M\), there exists a neighbourhood of \(x\) which doesn’t contain any points of \(M\), i.e. \(x\) has a neighbourhood contained in \(T-M\). By the result of Problem 1, \(T-M\) is open, hence \(M\) is closed.
Conversely, if \(M\) is closed, then \(T-M\) is open. Let \(x \in T-M\) and assume for contradiction that \(x\) is a contact point of \(M\). Then every neighbourhood of \(x\) contains at least one point of \(M\). However, \(T-M\) is open, and hence a neighbourhood of \(x\), and it doesn’t include any points of \(M\). This is a contradiction. Therefore the only contact points of \(M\) are in \(M\), i.e. \(M = [M]\).
\([M]\) is the smallest closed set containing \(M\).
If \([M]\) is not the smallest closed set, then there exists \(x \in [M]\), a contact point of \(M\), such that \(x \notin M\) and \(x \notin M’\). As we assume for contradiction that \(M’ = [M] - \{x\}\) is closed, then \(T - M’ = T - ([M] - \{x\})\) is open, but \(x\) is a contact point of \(M\) so every neighbourhood contains an element of \(M\). This leads to a contradiction.
The closure operator satisfies Theorem 1 on page 46 (Metric Spaces: convergence. Open and closed sets), i.e.
If \(M \subseteq N\), then \([M] \subseteq [N]\);
Let \(M \subseteq N\). Assume for contradiction that there exists \(x \in [M]\) such that \(x \notin [N]\). But then \(x\) is a contact point of \(M\), and not of \(N\), which is impossible as \(M\) is contained in \(N\). Contradiction.
\([[M]]\) = [M];
\([M]\) is closed, so by the result of 2.1, \([[M]] = [M]\).
\([M \cup N] = [M] \cup [N]\);
\(M \cup N \subseteq [M] \cup [N]\). Taking closure of both sides, we have \([M \cup N] \subseteq [[M] \cup [N]] = [M] \cup [N]\), as the union of two closed sets is itself closed.
Both \([M] \subseteq [M \cup N]\) and \([N] \subseteq [M \cup N]\). Hence \([M] \cup [N] \subseteq [M \cup N]\), since both \([M]\) and \([N]\) are subsets.
Combining the two inequalities, we get the desired result.
\([\emptyset] = \emptyset\).
\(\emptyset\) is closed, therefore \([\emptyset] = \emptyset\).
Consider the set \(\mathscr{T}\) of all possible topologies defined in a set \(X\), where \(\tau_2 \leq \tau_1\) means \(\tau_2\) is weaker than \(\tau_1\). Verify \(\leq\) is a partial ordering of \(\mathscr{T}\). Does \(\mathscr{T}\) have maximal and minimal elements? If so, what are they?
We need to show that the relation is reflexive, transitive and antisymmetric.
Reflexive as any topology \(\tau\) contains all of its own sets, i.e. \(\tau \leq \tau\).
Transitive since if \(\tau_1 \leq \tau_2\), i.e. all of the sets in \(\tau_1\) are in \(\tau_2\), and \(\tau_2 \leq \tau_3\), i.e. all of the sets in \(\tau_2\) are in \(\tau_3\), then \(\tau_1 \leq \tau_3\).
Antisymmetry follows too, since if \(\tau_2\) contains all of the sets of \(\tau_1\), then \(\tau_2 \leq \tau_1\) only if \(\tau_1\) contains all of the sets of \(\tau_2\), i.e. only if they are equal. Hence \(\tau_1 \leq \tau_2\) and \(\tau_2 \leq \tau_1\) only occurs when \(\tau_1 = \tau_2\).
Not a full ordering. Take two topologies on \(X = \{1, 2\}\), with \(\tau_1 = \{\emptyset, \{1\}, X\}\) and \(\tau_2 = \{\emptyset, \{2\}, X\}\). Evidently \(\tau_1\) and \(\tau_2\) cannot be compared as \(\tau_1\) does not include \(\{2\}\) and \(\tau_2\) does not include \(\{1\}\).
Minimal topology is \(\{\emptyset, X\}\). Maximal topology is the topology where every subset of \(X\) is open.
Can two distinct topologies \(\tau_1\) and \(\tau_2\) in \(X\) generate the same relative topology in a subset \(A \subseteq X\)?
Yes, if they have the same subsets of \(A\) in the topology but different subsets of \(X - A\).
Let \(X = \{a, b, c\}\), \(A = \{a, b\}\), \(B = \{b, c\}\) and \(\mathscr{G} = \{\emptyset, X, A, B\}\). Is \(\mathscr{G}\) a base for a topology in \(X\)?
No, since \(A \cap B = \{b\}\) is open but cannot be described as a union of elements of the base.
Prove if \(M\) is an uncountable subset of a topological space with a countable base, then some point of \(M\) is a limit point of \(M\).
Assume for contradiction that \(M\) doesn’t have a limit point. Then for any \(x\in M\) there exists some \(G_x\in\mathscr{G}\) such that \(G_x\cap M = \{x\}\). For another point \(y\in M\) we can also find \(G_y \cap M = \{y\}\). Hence \(x \not= y \Rightarrow G_x \not= G_y\), which defines an injection from \(M\) to \(\mathscr{G}\), so \(|M|\leq|\mathscr{G}|\). Hence \(M\) is countable, but this is a contradiction. Therefore there must be a limit point in \(M\). [Credit to CoveredInChocolate for this proof.]
Prove \(T\) in Example 4, from page 79, (where \(T = \{a, b\}\) and the open sets are \(\emptyset\), \(T\), and \(\{b\}\)) is connected.
Comment. \(T\) might be called a ‘connected doubleton’.
\(\{b\}\) is not closed as \(T - \{b\} = \{a\}\) is not open. Hence \(\emptyset\) and \(T\) are the only sets which are both open and closed, therefore \(T\) is connected.
Prove a topological space satisfying the second axiom of countability automatically satisfies the first axiom of countability.
If \(T\) satisfies the second axiom of countability, i.e. \(T\) has a countable base \(\mathscr{G}\), then by Theorem 3, for any \(O_x \subseteq T\) containing \(x\), there is a set \(G_\alpha \in \mathscr{G}\) such that \(x \in G_\alpha \subseteq O_x\), i.e. the topology satisfies the first axiom of countability.
Given an example of a topological space satisfying the first axiom but not the second.
A discrete, uncountable topology satisfies the first axiom of countability, but not the second axiom. In the discrete topology, every distinct point is an open set, so it cannot be covered by a countable base. [Credit to CoveredInChocolate for this example.]
Let \(\tau\) be the system of sets containing the empty set and every subset of the closed unit interval \([0, 1]\) obtained by deleting a finite or countable number of points from \(X\). Verify that \(T = (X, \tau)\) is a topological space. Prove that \(T\) satisfies neither the second nor the first axiom of countability. Prove \(T\) is a \(T_1\)-space, but not a Hausdorff space.
\(\emptyset\) in \(\tau\) and \(X\) in \(\tau\) for the cases where no points are deleted. Need to show open sets are closed under arbitrary unions and finite intersections.
Let \(G_\alpha\) be the open sets in τ and \(F_\alpha\) be the corresponding closed sets. Then for any arbitrary index set \(I\), \[ \cup_\alpha G_\alpha = \cup_\alpha (X - F_\alpha) = X - \cap_\alpha F_\alpha. \] Since closed sets are closed under arbitrary intersections, and \(\cap_\alpha F_\alpha \subseteq F_\beta\) for some closed set which is at most countable, the arbitrary union is \(X\) with a countable number of points removed. So closed under arbitrary unions.
For finite \(k\), \(\cap_k G_k = \cap_k (X - F_k) = X - \cup F_k\). A finite union of closed sets is closed, and a finite union of countable sets is countable, hence the topology is closed under finite intersections.
By Problem 8, for the countability axioms, we simply need to show that the space does not satisfy the first axiom of countability. Let \(G\) be an open set of \(x\) where all of the rationals (except \(x\) if it is rational) have been removed. Then there is only one open set \(O\) such that \(x \in O \subseteq G\), namely, \(G\). So the first axiom is not satisifed, and hence neither is the second. [Not sure whether this argument is sound, since one open set is still countable.]
Let \(x\) and \(y\) be two distinct points and \(O_x = [0, y) \cup (y, 1]\) and \(O_y = [0, x) \cup (x, 1]\). Then \(x \notin O_y\) and \(y \notin O_x\), hence \(T_1\)-space.
Any open set including \(x\) will only have a countable set of \(X\) removed, as will any open set including \(y\). Hence \(O_x \cap O_y \not= \emptyset\), so not Hausdorff.
Let \(T\) be the same as in Problem 10. Prove that the only convegent sequences in \(T\) are the ‘stationary sequences’, i.e. the sequences all of whose terms are the same starting from some index \(n\). Prove that the set \(M = (0, 1]\) has the point \(0\) as a contact point, but contains no sequence of points converging to \(0\).
For contradiction, assume that there is a non-stationary sequence, i.e. (without loss of generalisation) for all \(x_n\) in the sequence, \(O_{x_{n+1}} \subseteq O_{x_{n}}\) and \(x_{n+1} \not= x_n\). However, this defines a countable neighbourhood base of the limit point \(x\), but the space does not satsify the first axiom of countability, so there is a contradiction.
\(0\) is a contact point of \(M\), as for any open set \(O_0\) of \(0\) \(O_0 \cap M \not= \emptyset\), remember (\(\{0\}\) is not an open set).
Since the only convergent sequences in \(T\) are ‘stationary sequences’, no sequence of points in \(M = (0, 1]\) can converge to a point outside of \(M\), so no sequence can converge to \(0\).
Prove the converse of Theorem 8, i.e. prove if every finite subset of \(T\) is closed, \(T\) is a \(T_1\)-space.
Comment. Hence a topological space \(T\) is a \(T_1\)-space iff every finite subset of \(T\) is closed.
If every finite subset of \(T\) is closed, then for all \(x \in T\), \(\{x\}\) is closed, i.e. \(T - \{x\}\) is open. Then for any two distinct points \(x\) and \(y\), we can construct two neighbourhoods \(x \in O_x = T - \{y\}\) and \(y \in O_y = T - \{x\}\) such that \(x \notin O_y\) and \(y \notin O_x\). Hence \(T\) is a \(T_1\)-space.
Prove the following theorem, known as Urysohn’s lemma: given a normal space \(T\) and two disjoint closed sets \(F_1, F_2 \in T\), there exists a continuous real function \(f\) such that \(0 \leq f(x) \leq 1\) and \[f(x) = \begin{cases} 0 \textrm{ if } x \in F_1 \\ 1 \textrm{ if } x \in F_2. \end{cases}\]
\(f\) is a mapping from \(T\) to \([0, 1]\), which is a metric space and hence a normal topological space (and also therefore a \(T_1\)-space). The preimage of \(\{0\}\), which is closed in \([0, 1]\) (by Theorem 8), is \(F_1\), which is closed in \(T\), and the preimage of \(\{1\}\), which too is closed in \([0, 1]\), is \(F_2\), which is closed in \(T\).
[TODO: Unclear what steps then need to be taken here…]
Maybe this video by Daniel Murfet for the class MAST30026 is useful.
A \(T_1\)-space \(T\) is said to be completely regular if, given any closed set \(F \subseteq T\) and any point \(x_0 \in T - F\), there exists a continuous real function \(f\) such that \(0 \leq f(x) \leq 1\) and \[f(x) = \begin{cases} 0 \textrm{ if } x = x_0 \\ 1 \textrm{ if } x \in F. \end{cases}\] (Completely regular spaces are also called Tychonoff spaces.) Prove that every normal space is completely regular, but not conversely. Prove that every subspace of a completely regular space (in particular, of a normal space) is completely regular.
TODO.
