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MATH31052 Topology

2 Topological Spaces

  • 2.1 Problem.   What properties of a subset of Euclidean space are preserved by a homeomorphism or topological equivalence, in other words determine `the topology'?

  • 2.2 Definition.   Suppose \(X\) is a subset of a Euclidean space, \(\x _0\in X\) and \(\e >0\). Then the open \(\e \)-ball about \(\x _0\) (in \(X\)) is the set

    \[B_{\e }^X(\x _0) = \{\,\x \in X\mid |\x -\x _0|<\e \,\}.\]

    For \(X=\R ^n\), this is often written \(B_{\e }(\x _0)\) so that, for \(X\subset \R ^n\), \(B_{\e }^X(\x _0) = B_{\e }(\x _0)\cap X\).

  • 2.3 Remark.   \(B^X_\e (\x _0)\) in general depends on \(X\), e.g. \(B_1^{\R }(0) = (-1,1)\), \(B_1^{[0,\infty )}(0) = [0,1)\), \(B_1^{\Z }(0) = \{0\}\).

  • 2.4 Proposition.   A function \(f\colon X \rightarrow Y\) of subsets of Euclidean spaces is continuous at \(\x _0\in X\) if and only if, for each real number \(\e >0\), there exists real number \(\delta >0\) such that

    \begin{equation} \x \in B_{\delta }^X(\x _0) \Rightarrow f(\x ) \in B_{\e }^Y\bigl (f(\x _0)\bigr ) \end{equation}

    or (equivalently)

    \begin{equation} B_{\delta }^X(\x _0) \subset f^{-1}\Bigl (B_{\e }^Y\bigl (f(\x _0)\bigr )\Bigr ). \end{equation}

  • Proof. (1) is a restatement of Definition 0.21 and then (2) is a rewrite of (1) using Definition 0.13(b).  □ class="theoremendmark" >

  • 2.5 Definition.   Suppose that \(X\) is a subset of a Euclidean space. A subset \(U\) of \(X\) is a neighbourhood of a point \(\x _0\in X\) if there exists a real number \(\e >0\) such that \(B^X_{\e }(\x _0)\subset U\).

    A subset \(U\subset X\) is open (in \(X\)) if it is a neighbourhood of each point \(\x _0\in U\).

  • 2.6 Example.  

    • (a) For any subset \(X\) of a Euclidean space, \(X\) is open in \(X\) using any \(\e >0\) since by definition \(B_{\e }^X(\x _0)\subset X\).

    • (b) For any subset \(X\) of a Euclidean space, the empty set \(\emptyset \) is open in \(X\) since the condition is vacuous.

    • (c) An open interval \((a,b)\) is open in \(\R \) [which means that the language is consistent] since for \(x_0\in (a,b)\) we can put \(\e = \min (x_0-a,b-x_0)\).

    • (d) The singleton set \(\{0\}\) is not open in \(\R \) (since \(B_{\e }(0)\) contains \(\e /2 \not \in \{0\}\) for all \(\e >0\)) but it is open in \(\Z \) (since, taking \(\e =1\), \(B_1^{\Z }(0)=\{0\} \subset \{0\})\).

  • 2.7 Proposition.   For \(\x _0\in X\subset \R ^n\) and \(\e >0\), the open \(\e \)-ball \(B_{\e }^X(\x _0)\) is open in \(X\).

  • Proof. Given \(\x _1\in B_{\e }^X(\x _0)\), then \(|\x _0-\x _1|<\e \). Put \(\e _1 = \e - |\x _0-\x _1|>0\). Then \(B_{\e _1}^X(\x _1) \subset B_{\e }^X(\x _0)\) as required to prove that \(B_{\e }(\x )\) is open [for \(\x \in B_{\e _1}^X(\x _1) \Rightarrow |\x - \x _1| < \e _1 = \e - |\x _0 - \x _1| \Rightarrow |\x - \x _1| + |\x _1 - \x _0| < \e \Rightarrow |\x -\x _0|<\e \) (since \(|\x -\x _0| \leq |\x - \x _1| + |\x _1 - \x _0|\) by the Triangle Inequality) \(\Rightarrow \x \in B_{\e }^X(\x _0)\).]  □

  • 2.8 Theorem.   A function \(f\colon X \rightarrow Y\) of subsets of Euclidean spaces is continuous if and only if, for each open subset \(V\) of \(Y\), the subset \(f^{-1}(V)\) is open in \(X\).

  • Proof. `\(\Rightarrow \)': Suppose that \(f\colon X\to Y\) is continuous. Let \(V\) be an open subset of \(Y\). Then, to see that \(f^{-1}(V)\) is open in \(X\), let \(\x _0\inf ^{-1}(V)\). Then \(f(\x _0)\in V\) and so, since \(V\) is open in \(Y\), there exists \(\e >0\) such that \(B_{\e }^Y\bigl (f(\x _0)\bigr ) \subset V\). So \(f^{-1}\bigl (B_{\e }^Y(f(\x _0))\bigr ) \subset f^{-1}(V)\). Now, since \(f\) is continuous at \(\x _0\), by Proposition 2.4 there exists \(\delta >0\) such that \(B_{\delta }^X(\x _0) \subset f^{-1}\bigl (B_{\e }^Y(f(\x _0))\bigr )\). Thus \(B_{\delta }^X(\x _0) \subset f^{-1}(V)\) as required to prove that \(f^{-1}(V)\) is a neighbourhood of \(\x _0\). Hence \(f^{-1}(V)\) is open in \(X\).

    `\(\Leftarrow \)': Suppose that the condition in the Theorem holds and \(\x _0\in X\). To see that \(f\) is continuous at \(\x _0\) suppose that \(\e >0\). Then, by Proposition 2.7, \(B_{\e }^Y\bigl (f(\x _0)\bigr )\) is open in \(Y\). Hence, by hypothesis, \(f^{-1}\bigl (B_{\e }^Y(f(\x _0))\bigr )\) is open in \(X\). So, since \(\x _0 \in f^{-1}\bigl (B_{\e }^Y(f(\x _0))\bigr )\), there exists \(\delta >0\) such that \(B_{\delta }^X(\x _0) \subset f^{-1}\bigl (B_{\e }^Y(f(\x _0))\bigr )\) and so, by Proposition 2.4, \(f\) is continuous at \(\x _0\). Hence \(f\colon X\to Y\) is continuous.  □

  • 2.9 Corollary.   A bijection \(f\colon X \rightarrow Y\) of subsets of Euclidean spaces is a homeomorphism if and only if

    \[U \mbox { open in }X \Leftrightarrow f(U) \mbox { open in }Y.\]

  • Proof. Suppose that \(f\colon X\to Y\) is a bijection of subsets of Euclidean space.

    (a) For each \(V\subset Y\), \(V=f(U)\) where \(U=f^{-1}(V)\) (since \(f\) is a bijection). Hence \(f\) is continuous if and only if (\(V\) open in \(Y \Rightarrow f^{-1}(V)\) open in \(X\)) (Theorem 2.8) if and only if (\(f(U)\) open in \(Y \Rightarrow U\) open in \(X\)).

    (b) Let \(\g =f^{-1}\colon Y\to X\). Then \(\g ^{-1}=f\). Hence, for \(U\subset X\), \(\g ^{-1}(U)=f(U)\). So \(\g =f^{-1}\) is continuous if and only if (\(U\) open in \(X \Rightarrow \g ^{-1}(U)\) open in \(Y\)) if and only if (\(U\) open in \(X \Rightarrow f(U)\) open in \(Y\)). □

  • 2.10 Remark.   This result indicates that the answer to Problem 2.1 is that a homeomorphism is a bijection which preserves the open sets and so `the topology' is determined by the open sets. So we can specify a `topology' on any set \(X\) by declaring which subsets of \(X\) are to be the open sets.

  • 2.11 Definition.   Given a set \(X\), a topology on \(X\) is a collection \(\tau \) of subsets of \(X\) with the following properties:

    • (i) \(\emptyset \in \tau \), \(X\in \tau \);

    • (ii) the intersection of any two subsets in \(\tau \) is in \(\tau \):

      \[U_1,U_2\in \tau \Rightarrow U_1\cap U_2\in \tau ;\]

    • (iii) the union of any collection of subsets in \(\tau \) is in \(\tau \):

      \[U_{\lambda }\in \tau \mbox { for all }\lambda \in \Lambda \Rightarrow \bigcup _{\lambda \in \Lambda }U_{\lambda }\in \tau .\]

    A pair \((X,\tau )\) such that \(\tau \) is a topology on \(X\) is called a topological space, usually denoted \(X\) when the topology is clear. The subsets in \(\tau \) are called the open subsets of \(X\) (with the topology \(\tau \)) or the open sets of the topology \(\tau \). Thus, given a topology \(\tau \) on a set \(X\) the statements `\(U\in \tau \)' and `\(U\) is a open subset in \(X\)' have precisely the same meaning.

  • 2.12 Definition.   Suppose that \((X,\tau _1)\) and \((Y,\tau _2)\) are topological spaces. A function \(f\colon X\rightarrow Y\) is continuous (with respect to \(\tau _1\) and \(\tau _2\)) if, for each open subset \(V\) of \(Y\) (with the topology \(\tau _2\)), the subset \(f^{-1}(V)\) is open in \(X\) (with the topology \(\tau _1\)), i.e.

    \[V\in \tau _2\Rightarrow f^{-1}(V)\in \tau _1.\]

    The function \(f\colon X\rightarrow Y\) is a homeomorphism if it is a continuous bijection with continuous inverse.

  • 2.13 Proposition.   If \(f\colon X\rightarrow Y\) and \(g\colon Y\rightarrow Z\) are continuous functions of topological spaces, then \(g\circ f\colon X\rightarrow Z\) is a continuous function.

  • Proof. Exercise.  □ class="theoremendmark" >

  • 2.14 Proposition.   A bijection \(f\colon X\rightarrow Y\) between topological spaces is a homeomorphism if and only if

    \[\mbox {$U$ is open in $X$} \Leftrightarrow \mbox {$f(U)$ is open in $Y$}.\]

  • Proof. Exercise. (This is identical to the proof of Corollary 2.9 with Definition 2.12 playing the role of Theorem 2.8.)  □ class="theoremendmark" >

  • 2.15 Remark.   The original definition of topological space given by Felix Hausdorff in 1914 was equivalent to Definition 2.11 with the addition of a fourth condition on the set of open sets (now called the Hausdorff condition to be discussed in Ÿ4).

  • 2.16 Proposition.  

    • (a) Suppose that \(X\) is a subset of Euclidean space. Then the open subsets of \(X\) defined by Definition 2.5 are the open subsets of a topology on \(X\) (called the usual topology on \(X\)).

    • (b) A function \(f\colon X \rightarrow Y\) of subsets of Euclidean spaces is continuous according to Definition 0.21 if and only if it is continuous according to Definition 2.12 with respect to the usual topologies on \(X\) and \(Y\).

  • Proof. (a) To see that the open sets of a subset of Euclidean space \(X\) given by Defintion 2.5 define a topology on \(X\) we need to check the conditions in Definition 2.11.

    (i) \(\emptyset \) and \(X\) are open in \(X\) by Examples 2.6(a) and (b).

    (ii) Suppose that \(U_1\subset X\) and \(U_2\subset X\) are open in \(X\). Then \(U_1\cap U_2\) is open in \(X\) (Exercise).

    (iii) Suppose that \(U_{\lambda }\subset X\) is open in \(X\) for all \(\lambda \in \Lambda \). Then \(\bigcup _{\lambda \in \Lambda }U_{\lambda }\) is open in \(X\) (Exercise).

    Hence these open sets are the open sets of a topology on \(X\).

    (b) This is a restatement of Theorem 2.8.  □

  • 2.17 Example.  

    • (a) The discrete topology on a set \(X\) consists of all the subsets of \(X\).

      If \(X\) has the discrete topology and \(Y\) is any topological space, then all functions \(f\colon X\rightarrow Y\) are continuous.

    • (b) The indiscrete topology on a set \(X\) is given by \(\tau =\{\emptyset ,X\}\).

      If \(X\) has the indiscrete topology and \(Y\) is any topological space, then all functions \(f\colon Y\rightarrow X\) are continuous.

    • (c) Suppose that \(X=\{a,b\}\), a two point set. Then there are four topologies on \(X\):

      • (i) \(\tau _1=\{\emptyset ,X\}\) (the indiscrete topology);

      • (ii) \(\tau _2=\{\emptyset ,\{a\},X\}\);

      • (iii) \(\tau _3=\{\emptyset ,\{b\},X\}\);

      • (iv) \(\tau _2=\{\emptyset ,\{a\},\{b\},X\}\) (the discrete topology).

      The topological spaces \((X,\tau _2)\) and \((X,\tau _3)\) are homeomorphic with a homeomorphism given by \(a\mapsto b\), \(b\mapsto a\). Either of these topologies on a set of two elements is called the Sierpinski topology.

    • (d) Suppose that \(X=\{a,b,c\}\) is a three point set.

      • (i) \(\{\emptyset ,\{a,b\},\{b,c\},X\}\) is not a topology (intersection fails).

      • (ii) \(\{\emptyset ,\{a\},\{b\},X\}\) is not a topology (union fails).

    • (e) The identity function \(\id _X\colon X\rightarrow X\), given by \(\id _X(x)=x\) for all \(x\in X\), is continuous with respect to any topology on the set \(X\) (Exercise).

    • (f) Given sets \(X\) and \(Y\) and a point \(a\in Y\), the constant function \(c_a\colon X\rightarrow Y\), given by \(c_a(x)=a\) for all \(x\in X\), is continous with respect to any topologies on \(X\) and \(Y\) (Exercise).

  • 2.18 Remark.  

    • (a) By induction, the intersection of any finite number of open subsets in a topology is open. However, we do not require the intersection every collection of open subsets to be open. For example, in \(\R \) with the usual topology, \(\bigcap _{n=1}^{\infty }(-1/n,1/n) = \{0\}\) which is not open.

    • (b) Continuity does not imply that the image of each open set if open. For example, in \(\R \) with the usual topology the continuous map \(f\colon \R \to \R \) given by \(f(x) = x^2\) has \(f(-1,1) = [0,1)\) which is not open.

      A continuous function \(f\colon X\to Y\) of topological spaces for which \(U\) open in \(X\) implies that \(f(U)\) is open in \(Y\) is called an open map.

  • 2.19 Definition.   A subset \(A\) of a topological space \(X\) is closed when its complement \(X\setminus A\) is open.

  • 2.20 Proposition.   In a topological space \(X\),

    • (i) \(\emptyset \) and \(X\) are closed;

    • (ii) the union of any pair of closed subsets is closed;

    • (iii) the intersection of any collection of closed subsets is closed.

    Furthermore, any collection of subsets of \(X\) satisfying these conditions is the set of closed subsets of a topology on \(X\).

  • Proof. This is immediate from Definition 2.11 using the set theoretic properties of complements of unions and intersections (the de Morgan Laws, Proposition 0.5(iv)). class="theoremendmark" > □

  • 2.21 Remark.   The word `closed' is not the same as `not open'. For example, in \(\R \) with the usual topology, \([0,1)\) is neither open nor closed whereas \(\emptyset \) is both open and closed. In general, most subsets are neither open nor closed. Some subsets are both open and closed.

  • 2.22 Definition.   Suppose that \(X\) is a topological space. Then a collection \(\B \) of open subsets in \(X\) is called a basis for the topology on \(X\) if every non-empty open subset in \(X\) can be expressed as a union of open subsets in \(\B \).

  • 2.23 Proposition.   A basis for the usual topology on \(\R \) is given by

    \[\B =\{\,(a,b)\mid a,b\in \R ,\, a<b\,\}.\]

  • Proof. Given an open set \(U\subset \R \) (in the usual topology) then, for each \(x\in U\) there exists \(\e _x>0\) such that \(B_{\e _x}(x) = (x - \e _x, x+ \e _x) \subset U\). Then \(U = \bigcup _{x\in U} (x - \e _x,x+\e _x)\) (the empty union if \(U=\emptyset \)).  □

  • 2.24 Remark.   The material in Ÿ1 about paths, path-connectedness and cut-points extends from subsets of Euclidean spaces to all topological spaces.

  • 2.25 Definition.   For a subset \(A \subset X\) of a topological space \(X\) the interior \(A^\circ \) is defined to be the union of all open subsets \(U\) of \(X\), which are contained in \(A\).

    The closure \(\overline {A}\) is defined as the intersection of all closed subsets of \(X\) which contain \(A\).

    The boundary \(\partial A\) is defined as the difference \(\overline {A} \setminus A^\circ \).

  • 2.26 Definition.   An open neighbourhood of a point \(x\) in a topological space \(X\) is an open subset which contains \(x\).

  • 2.27 Example.   The consider the interval \(J=[0,1)\) as a subset of \(\R \) with the usual topology. Then we have \(J^\circ =(0,1)\) and \(\overline {J}=[0,1]\) and \(\partial J = \{0,1\}\).

    Indeed, \((0,1)\) is an open subset, which is contained in \(J\). Hence, \((0,1) \subset J^\circ \), but every open subset which contains \(0\), will also contain the interval \((-\epsilon ,\epsilon ) \not \subset J\) for \(\epsilon > 0\) sufficiently small. Hence, \(0\) is not cointained in the interior.

    Similarly \([0,1]\) is a closed subset, which contains \(J\). Hence, \([0,1) \subset \overline {J} \subset [0,1]\). Hence, \(\overline J\) has to be either \([0,1)\) or \([0,1]\). Since the closure of a set is a closed by definition. it follows, that \(\overline J=[0,1]\).

    Now, we obtain \(\partial J = [0,1]\setminus (0,1) = \{0,1\}\).

Exercises
  • 1 .  (a) Prove that \((0,1)\) is open in \(\R \).

    (b) Prove that \((0,1]\) is open in \([-1,1]\) but not in \(\R \).

    (c) Prove that \(\{1\}\) is open in \(\Z \), the subset of integers, but not in \(\R \).

  • 2 .  Which of the following collections of subsets, together with the empty set and \(\R \) are a topology on \(\R \)?

    • (a) the finite subsets of \(\R \) (not the finite intervals);

    • (b) the subsets of \(\R \) whose complements are finite;

    • (c) all subsets of the form \((a,\infty )=\{\,x\in \R \mid x>a\,\}\);

    • (d) all subsets of the form \([a,\infty )=\{\,x\in \R \mid x\geq a\,\}\);

    • (e) all subsets \(U\subset \R \) such that \(0\in U\);

    • (f) all subsets \(U\subset \R \) such that \(0\not \in U\).

  • 3 .  Find all topologies on a set of three elements (say \(X=\{a,b,c\}\)) and divide them into homeomorphism classes.

  • 4 .  Prove that, if \(f\colon X\rightarrow Y\) and \(g\colon Y\rightarrow Z\) are continuous functions of topological spaces, then \(g\circ f\colon X\rightarrow Z\) is a continuous function. [Proposition 2.13]

  • 5 .  Prove that a bijection \(f\colon X\to Y\) between topological spaces is a homeomorphism if and only if

    \[U\mbox { is open in }X \Leftrightarrow f(U)\mbox { is open in }Y.\]

    [Proposition 2.14]

  • 6 .  Prove that, for a topological space \(X\), the identity function \(I_X\colon X\to X\) given by \(I_X(x)=x\) for all \(x\in X\) is a homeomorphism. [Example 2.17(c)]

  • 7 .  Prove that, for topological spaces \(X\) and \(Y\) and a point \(a\in Y\), the constant function \(c_a\colon X\to Y\) given bt \(c_a(x) = a\) for all \(x\in X\) is continuous. [Example 2.17(d)]

  • 8 .  Prove that, if \(X\) and \(Y\) are topological spaces, then \(f\colon X\rightarrow Y\) is continuous if and only if

    \[A \mbox { closed in }Y \Rightarrow f^{-1}(A) \mbox { closed in }X.\]

  • 9 .  Suppose that \(X\) and \(Y\) are topological spaces and \(\B \) is a basis for the topology of \(Y\). Prove that \(f\colon X\rightarrow Y\) is continuous if and only if

    \[V\in \B \Rightarrow f^{-1}(V)\mbox { is open in }X.\]

  • 10 .  Prove that a basis for the usual topology on \(\R ^n\) is provided by the set of all \(\e \)-balls, \(\B = \{\,B_{\e }(\x )\mid \x \in \R ^n\mbox { and }\e >0\,\}\).

  • 11 .  Prove that a collection of subsets \(\B \subset \P (X)\) of a set \(X\) is a basis for some topology on \(X\) if and only if

    • (a) for each \(x\in X\), there is a subset \(U\in \B \) such that \(x\in U\),

    • (b) given \(U_1\), \(U_2\in \B \) and \(x\in U_1\cap U_2\), there exists \(U\in \B \) such that \(x\in U\subset U_1\cap U_2\).

  • 12 .  Prove that a topological space \(X\) has a proper subset \(U\) (i.e. \(U\not =\emptyset \) and \(U\not =X\)) which is both open and closed if and only if there is a continuous surjection \(X\rightarrow \{0,1\}\) (where \(\{0,1\}\) has the usual topology). Prove that such a topological space is not path-connected.

  • 13 .  Let \(X = \{a,b\}\). For which topologies of Examples 2.17(c) is \(X\) path-connected?

  • 14 .  Given a commutative \(R\) ring with \(1\). The set of it's prime ideals (an ideal \(\mathfrak p \subsetneq R\) is called prime if for \(fg \in \mathfrak p\) it follows that either \(f \in \mathfrak p\) or \(g \in \mathfrak p\)) is called the spectrum of \(R\) and it is denoted by \(\text {Spec}(R)\). For an ideal \(I\) of \(R\) one considers the subset \(U_I = \{\mathfrak p \in \text {Spec}(R) \mid I \not \subseteq \mathfrak p\}\). Show that the subset of the form \(U_I\) form a topology for \(\text {Spec}(R)\).

  • 15 .  A point \(x\) is contained in the interior of \(A \subset X\) if and only if there exists an open neigbourhood \(U\) of \(x\), which is contained in \(A\).

    A point \(x \in X\) is contained in \(\overline A\) if and only if every open neighbourhood of \(x\) intersects \(A\).

    A point \(x \in X\) is contained in \(\partial A\) if and only if every open neighbourhood of \(x\) intersects \(A\) and its complement \(X \setminus A\).