Topology: General Topology 2021

Let's illustrate this last point with an interesting example. Let \(\PP^1_{\RR}\) be the set of \(1\)-dimensional linear subspaces of \(\RR^2\text{.}\) In other words, an element of \(\PP^1_{\RR}\) is a line in \(\RR^2\) through the origin. If \(L \in \PP^1_{\RR}\text{,}\) let \(\mu(L)\) denote the slope of \(L\text{;}\) that is, if \(L\) is spanned by a nonzero vector \((x,y) \in \RR^2\text{,}\) then

We will say that a line \(L \in \PP^1_{\RR}\) is close to a subset \(S \subseteq \PP^1_{\RR}\) if and only if one of the following conditions holds:

• if \(\mu(L) \neq \infty\text{,}\) then for every \(\varepsilon>0\text{,}\) there exists a line \(L' \in S\text{,}\) such that

  \begin{equation*} |\mu(L) - \mu(L')| \lt \varepsilon\text{,} \end{equation*}

  or

• if \(\mu(L) = \infty\text{,}\) then for every real number \(N\text{,}\) there exists a line \(L' \in S\) such that

  \begin{equation*} |\mu(L')|>N\text{.} \end{equation*}

Equivalently, if we think of the circle \(S^1 \subset \RR^2\text{,}\) then every line \(L \in \PP^1_{\RR}\) intersects \(S^1\) in exactly two points

\(i(L)\) and \(j(L)\text{,}\) where \(i(L) = -j(L)\text{.}\) We may now say that \(L \in \PP^1_{\RR}\) is close to \(S \subseteq \PP^1_{\RR}\) if and only if, for every \(\varepsilon>0\text{,}\) there exists a line \(L' \in S\) such that either \(\|i(L)-i(L')\|\lt \varepsilon\) or \(\|i(L)-j(L)\|\lt \varepsilon\text{.}\)

We have already seen the first example. Any subset \(X \subseteq \RR^n\) inherits a topology \(\tau_X\) that carries a subset \(S \subseteq X\) to

When \(X=\RR^n\text{,}\) this is often called the standard topology, and for \(X \subseteq \RR^n\text{,}\) this is called the subspace topology.

The closure operation on \(\PP^1_{\RR}\) (Example 2.1.1) is a topology.

1. Let \(S \subseteq \PP^1_{\RR}\text{.}\) If \(L \in S\text{,}\) then the conditions are automatic for \(L \in \tau(S)\text{.}\)

2. Let \(S_1, \dots, S_n \subseteq \PP^1_{\RR}\) be a collection of subsets of \(\PP^1_{\RR}\text{,}\) and let \(L \in \PP^1_{\RR}\text{.}\) If \(L\) is close to \(S_1 \cup \cdots \cup S_n\text{,}\) then a Pigeonhole argument like the one given in Proposition 1.1.4 ensures that \(L\) is close to \(S_1 \cup \cdots \cup S_n\text{.}\)

   If \(L\) is close to some \(S_i\text{,}\) then it is certainly close to \(S_1 \cup \cdots \cup S_n\text{.}\)

3. Let \(S \subseteq \PP^1_{\RR}\text{,}\) and let \(L \in \PP^1_{\RR}\) be a line that is close to \(\tau(S)\text{.}\) We aim to show that \(L\) is close to \(S\text{;}\) so let \(\varepsilon>0\text{.}\) There exists a line \(L' \in \tau(S)\) such that either \(\|i(L) - i(L')\|\lt \varepsilon\) or \(\|i(L)-j(L')\|\lt \varepsilon/2\text{.}\) Accordingly, there exists a line \(L'' \in S\) such that either \(\|i(L') - i(L'')\| \lt \varepsilon/2\) or \(\|i(L') - j(L'')\| \lt \varepsilon/2\text{.}\) Analyzing the four options and applying the triangle inequality, we find that either \(\|i(L) - i(L'')\|\lt \varepsilon\) or \(\|i(L) - j(L'') \|\lt \varepsilon\text{.}\)

The empty set \(\emptyset\) admits a unique topology. The powerset \(\PP( \emptyset )\) is a singleton \(\{ \emptyset \}\text{,}\) and the identity is the only map \(\{ \emptyset \} \to \{ \emptyset \}\text{.}\) This map is a topology. The only subset available — \(\emptyset\) itself — is both open and closed.

More generally, on any set \(X\text{,}\) the identity map

is always a topology. This is called the discrete topology on a set \(X\text{.}\)

In the discrete topology, a point \(x \in X\) is close to a subset \(S \subseteq X\) if and only if \(x \in S\text{.}\) Every subset of \(X\) is both open and closed in the discrete topology.

Let us study topologies on the singleton \(\{ x \}\text{.}\) The power set \(\PP( \{ x \} )\) is a two-element set \(\{ \emptyset, \{ x \} \}\text{.}\) There are four maps from this set to itself, but most of these are not topologies. In order for a map \(\tau\) to be a topology, the first condition demands that \(\tau( \{ x \} ) = \{ x \}\text{,}\) and the second condition demands that \(\tau ( \emptyset ) = \emptyset\text{.}\) Thus the only topology on the singleton \(\{ x \}\) is the discrete topology.

Let \(X \coloneq \{ x , y \}\text{.}\) In order to specify a topology \(\tau\) on \(X\text{,}\) it is really only necessary to specify the subsets \(\tau ( \{ x \} )\) and \(\tau ( \{ y \} )\text{.}\) The first condition states that these subsets must contain the sets \(\{ x \}\) and \(\{ y \}\text{,}\) respectively, and the second and third conditions become automatic. That gives us four topologies on \(\{ x, y \}\text{:}\)

• the discrete topology, in which \(\tau( \{ x \} ) = \{ x \}\) and \(\tau( \{ y \} ) = \{ y \}\text{;}\)

• the topology in which \(\tau( \{ x \} ) = \{ x \}\text{,}\) and \(\tau( \{ y \} ) = X\text{;}\)

• the topology in which \(\tau( \{ y \} ) = X\text{,}\) and \(\tau( \{ x \} ) = \{ x \}\text{;}\) and

• the topology in which \(\tau( \{ x \} ) = \tau( \{ y \} ) = X\text{.}\)

Let us describe the relation of closeness for each of these topologies. The only questions that aren't answered by the axioms of a topology is: “is \(x\) close to \(\{ y \}\text{?}\)” and “is \(y\) close to \(\{ x \}\text{?}\)” Let's answer these in each case:

• in the discrete topology, \(x\) is not close to \(\{ y \}\text{,}\) and \(y\) is not close to \(\{ x \}\text{;}\)

• in the second of these topologies, \(x\) is close to \(\{ y \}\text{,}\) but \(y\) is not close to \(\{ x \}\text{;}\)

• in the third topology, \(y\) is close to \(\{ x \}\text{,}\) but \(x\) is not close to \(\{ y \}\text{;}\) and

• finally, in the last of these topologies, \(x\) is close to \(\{ y \}\) and \(y\) is close to \(\{ x \}\text{.}\)

The asymmetry in the second and third topologies is strange from our natural idea of what “close” ought to mean. Nevertheless, these topologies are actually interesting examples; even rather bizarre ideas of closeness can be captured in topology!

Finally, for each of these examples, let us determine which sets are open and closed. In each case, the only questions that aren't answered by the axioms are: “is \(\{ x \}\) open? closed?” and “is \(\{ y \}\) open? closed?”

• in the discrete topology, both \(\{ x \}\) and \(\{ y \}\) are both open and closed;

• in the second topology, \(\{ x \}\) is closed and not open, and \(\{ y \}\) is open and not closed.

• in the third topology, the situation is reversed: \(\{ x \}\) is open and not closed, and \(\{ y \}\) is closed and not open;

• in the last of these topologies, neither \(\{ x \}\) nor \(\{ y \}\) is open or closed.

For any set \(X\text{,}\) we define a map

by the formula

This is a topology called the indiscrete topology. Thus in the indiscrete topology, any point is close to any nonempty subset. The only subsets of \(X\) that are open or closed in the indiscrete topology are \(\emptyset\) and \(X\) itself.

Let \(P\) be a preorder. That is, \(P\) is a set equipped with a relation \(\leq\) such that:

• for every \(a \in P\text{,}\) one has \(a \leq a\text{;}\) and

• for every \(a,b,c \in P\text{,}\) if \(a \leq b\) and \(b \leq c\text{,}\) then \(a \leq c\text{.}\)

Then we can define a topology as follows. For any subset \(S \subseteq P\text{,}\) we define

In other words, \(a \in P\) is close to \(S\) if and only if some element of \(S\) exceeds it.

Let us quickly confirm that this is a topology:

1. Let \(S \subseteq P\) be a subset. Since \(s \leq s\) for every \(s \in S\text{,}\) it follows that \(S \subseteq \tau_{\leq}(S)\text{.}\)

2. Let \(S_1, \dots, S_n \subseteq P\) be a finite collection of subsets of \(P\text{.}\) Then

   \begin{align*} \tau_{\leq}(S_1 \cup \cdots \cup S_n) \amp = \{a\in P : (\exists s \in S_1 \cup \cdots \cup S_n)(a \leq s) \}\\ \amp = \{a \in P : (\exists i)(\exists s \in S_i)(a \leq s)\}\\ \amp = \tau_{\leq}(S)_1 \cup \cdots \cup \tau_{\leq}(S)_n \text{.} \end{align*}

3. Let \(S \subseteq P\) be a subset. Then

   \begin{align*} \tau_{\leq}(\tau_{\leq}(S)) \amp = \{a \in P : (\exists t \in \overline{S})(a \leq t) \}\\ \amp = \{a \in P : (\exists s \in S)(\exists t \in X)(a \leq t \leq s) \}\\ \amp = \{a \in P : (\exists s \in S)(a \leq s) \}\\ \amp = \tau_{\leq}(S) \text{.} \end{align*}

This topology is called the Alexandroff topology.

The closed subsets are the subsets \(Z \subseteq P\) with the property that if \(a \in P\text{,}\) \(z \in Z\text{,}\) and \(a \leq z\text{,}\) then \(a \in Z\) as well. The open subsets are the subsets \(U \subseteq P\) with the property that if \(a \in P\text{,}\) \(u \in U\text{,}\) and \(a \geq u\text{,}\) then \(a \in U\) as well.

Form the set \(\RR\sqcup\{\infty\}\text{,}\) where \(\infty\) is just some symbol such that \(\infty\notin\RR\text{.}\) We already have a topology on \(\RR\text{;}\) let's doctor it a mite to make a topology on \(\RR\sqcup\{\infty\}\text{.}\) To do this, we have to understand how our notion of closeness has changed. So we have a point \(x\in \RR\sqcup\{\infty\}\) and a subset \(S\subseteq\RR\sqcup\{\infty\}\text{,}\) and we want to see if \(x\) is close to \(S\text{.}\) There are two cases:

• If \(x\in\RR\text{,}\) then we declare that \(x\) is close to \(S\) if and only if \(x\) is close to \(S\cap\RR\) (for the standard topology on \(\RR\)).

• We declare that \(\infty\) is close to \(S\) if and only if, for every \(N\in\RR_{\geq 0}\text{,}\) there is a point of \(S\) that does not lie

  in \([-N,N]\text{.}\)

Let's see that this is a topology. In effect, you check everything by checking the two cases (\(x\in\RR\) or \(x\notin\RR\)) separately. In the first case, you use the fact that \(\RR\) has a topology already, and in the second, you give a little argument.

• By definition if \(x\) lies in a subset \(S\subseteq\RR\sqcup\{\infty\}\text{,}\) then \(x\) is close to \(S\text{.}\)

• Let \(\{S_1,\dots,S_n\}\) be a finite collection of subsets of \(\RR\sqcup\{\infty\}\text{.}\) If \(x\in \RR\text{,}\) then since one has

  \begin{equation*} (S_1\cup\dots\cup S_n)\cap\RR=(S_1\cap\RR)\cup\dots\cup (S_n\cap\RR)\text{,} \end{equation*}

  it follows that \(x\) is close to \(S_1\cup\dots\cup S_n\) if and only if \(x\) is close to \((S_1\cap\RR)\cup\dots\cup (S_n\cap\RR)\) (for the standard topology on \(\RR\)), if and only if \(x\) is close to \(S_i \cap \RR\) for some \(i\) (for the topology on \(\RR\)), if and only if \(x\) is close to \(S_i\text{.}\) Now \(\infty\) is close to \(S_1\cup\dots\cup S_n\) if and only if, for any \(N\in\RR\text{,}\) there is a point of \(S_1\cup\dots\cup S_n\) that does not lie in \([-N,N]\text{.}\) This happens if and only if there is an \(i\) such that for any \(N\in\RR\text{,}\) there is

  a point of \(S_i\) that does not lie in \([-N,N]\text{.}\)

• Finally, suppose that every point of a subset \(S\subseteq\RR\sqcup\{\infty\}\) is close to a subset \(T\subseteq\RR\sqcup\{\infty\}\text{.}\) Now if \(x\) is close to \(S\text{,}\) there are two options. First, if \(x\in\RR\text{,}\) then \(x\) is close to \(S\cap\RR\) and hence to \(T\cap\RR\text{.}\) Otherwise, if \(x=\infty\text{,}\) then if \(\infty\in T\) or \(\infty\in S\text{,}\) then we're done. If not, then for every \(N\in\RR\text{,}\) there is a point \(s_N\in S\subseteq\RR\) such that \(|s_N|>N\text{;}\) this point \(s_N\) is close to \(T\text{,}\) so if we set \(\epsilon_N\coloneq|s_N|-N\text{,}\) then there exists a point \(t_N\in T\) such that \(|s_N-t_N|\lt \epsilon_N\text{.}\) Thus \(|t_N|>N\text{,}\) and so \(\infty\) is close to \(T\text{.}\)

Define a topology on \(\RR\sqcup\{-\infty,+\infty\}\) in which, for any point \(x\in\RR\sqcup\{-\infty,+\infty\}\) and any subset \(S\subseteq\RR\sqcup\{-\infty,+\infty\}\text{,}\)

• If \(x\in\RR\text{,}\) then \(x\) is close to \(S\) if and only if \(x\) is close to \(S\cap\RR\text{.}\)

• We declare that \(-\infty\) is close to \(S\) if and only if either \(-\infty\in S\) or for any \(N\in\RR_{\geq 0}\text{,}\) there is a point \(s\in S\cap\RR\) such that \(s\lt -N\text{.}\)

• We declare that \(+\infty\) is close to \(S\) if and only if either \(+\infty\in S\) or for any \(N\in\RR_{\geq 0}\text{,}\) there is a point \(s\in S\cap\RR\) such that \(s>N\text{.}\)

This is a topology.

Here's an extremely important example in modern mathematics. This one will be following us around throughout this text. Let \(X \subseteq \RR^m\) be any subspace, and let \(n \in \NN\text{.}\) We are about to define a topological space \(C_n(X)\) called the configuration space of \(n\) points in \(X\). As a set, \(C_n(X)\) is the set of subsets \(\{x_1,\dots,x_n\} \subseteq X\) of \(n\) distinct points:

To endow it with a topology, we need to define an auxiliary topological space.

Recall that we defined the subspace

We now pass to a further subspace:

Now there is a map \(q \colon E_m(X) \to C_m(X)\) defined by

We now declare that \(U \subseteq C_n(X)\) is open if and only if \(q^{-1}(U) \subseteq E_n(X)\) is open. This defines a system of open sets (and hence a topology), since the formation of the inverse image preserves unions and intersections.

Any set \(X\) can be given the cofinite topology, in which a subset \(Z\subseteq X\) is declared to be closed if and only if either \(Z\) is finite or \(Z=X\text{.}\) Thus in the cofinite topology,