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Section 1.1 Basic definitions

Our starting place is the study of the subspaces of Euclidean space.

In topology, we are not interested in the metric \(d\) itself but in what it can tell us about the nearness of points and subsets:

Definition 1.1.1.

Let \(S \subseteq \RR^n\) be a subset. A point \(x \in \RR^n\) will be said to be close to \(S\) if and only if, for every \(\varepsilon > 0\text{,}\) there exists an element \(s \in S\) such that \(d(x,s) \lt \varepsilon\text{.}\)

We may express this notion with the aid of an auxiliary quantity. Define the distance from \(x\) to \(S\) as

\begin{equation*} d(x,S) \coloneq \inf \{ d(x,s) : s \in S \}\text{.} \end{equation*}

Then \(x\) is close to \(S\) if and only if \(d(x,S) =0\text{.}\)

Let \(X \subseteq \RR^n\) be a subset such that \(S \subseteq X\text{.}\) We shall write \(\tau_X(S) \subseteq X\) for the set of \(x \in X\) that are close to \(S\text{:}\)

\begin{equation*} \tau_X(S) \coloneq \{ x \in X : (\forall \varepsilon >0)(\exists s \in S)(d(x,s)\lt \varepsilon) \} = \{x \in X : d(x,S) = 0\} \end{equation*}

This is called the closure of \(S\) in \(X\). In other words,

\begin{equation*} \tau_X(S) = X \cap \bigcap_{\varepsilon>0}\bigcup_{s\in S} B^n(s,\varepsilon)\text{.} \end{equation*}

We will refer to \(X\) along with the map

\(\tau_X \colon \PP(X) \to \PP(X)\) as the subspace \(X \subseteq \RR^n\text{.}\)

Every time you see a new definition in mathematics, you should begin immediately to think about as many examples as you can. You should also have a couple of counterexamples at hand, preferably at least one for each condition in a definition. Then, when you encounter a theorem with a proof, you can take your example and “run” the proof on your example, and you can run it on your counterexample to see where the conditions are really being used. This is one of the most important ways to understand how theorems really work.

In that spirit, let's dig into some examples.

  • If \(X = \varnothing\text{,}\) then this definition becomes a little boring: the only subset of \(\varnothing\) is \(\varnothing\) itself, and \(\tau_{\varnothing}(\varnothing) = \varnothing\text{.}\)

  • If \(X = \RR\) itself, then let's see what happens for various choices of \(S\text{:}\)

    • Let's consider various unit intervals:

      \begin{equation*} \tau_{\RR}(\left]0,1\right[) = \tau_{\RR}([0,1]) = \tau_{\RR}(\left]0,1\right]) = [0,1] \end{equation*}

    • If \(\QQ \subset \RR\) denotes the set of rational numbers, then \(\tau_{\RR}(\QQ) = \RR\text{.}\)

    • At the same time, \(\tau_{\RR}(\RR \smallsetminus \QQ) = \RR\text{.}\)

    • Confirm that \(\tau_{\RR}(\varnothing)=\varnothing\text{.}\)

  • If \(X \subseteq \RR^n\) is a subspace, then we have

    \begin{equation*} \tau_X(S) = \tau_{\RR^n}(S) \cap X\text{.} \end{equation*}

    This is an important point: the closure of \(S\) is done relative to the subspace \(X\). Let's see how the closure of \(S = \left]0,1\right[\) depends on which subspace \(X \subseteq \RR\) in which we are closing:

    • If \(X=\RR\text{,}\) then \(\tau_{\RR}(\left]0,1\right[) = [0,1]\text{.}\)

    • If \(X=\left]0,1\right[\text{,}\) then \(\tau_{\left]0,1\right[}(\left]0,1\right[) = \left]0,1\right[\text{.}\)

    • If \(X=\left]0,1\right]\text{,}\) then \(\tau_{\left]0,1\right]}(\left]0,1\right[) = \left]0,1\right]\text{.}\)

    • If \(X=\left]0,+\infty\right[\text{,}\) then \(\tau_{\left]0,+\infty\right[}(\left]0,1\right[) = \left]0,1\right]\text{.}\)

Let \(X \subseteq \RR\) be a subspace. Let \(S \subseteq X\) be a nonempty subset that is bounded above. The completeness property of \(\RR\) then ensures that \(S\) has a supremum \(\sup S\text{.}\) If \(\sup S \in X\text{,}\) then \(\sup S\) is close to \(S\) in \(X\text{;}\) that is,

\begin{equation*} \sup S \in \tau_X(S)\text{.} \end{equation*}

Let's prove this; let \(x\coloneq \sup S\text{.}\) Let \(\varepsilon>0\text{.}\) If \(\left]x-\varepsilon, x+\varepsilon\right[ \cap S = \varnothing\text{,}\) then \(x-\varepsilon\) is also an upper bound for \(S\text{,}\) contradicting the assumption that \(x\) is the least upper bound for \(S\text{.}\) Consequently, there exists an element \(s \in S\) such that \(d(x,s) \lt \varepsilon\text{.}\)

The same argument shows that if \(S \subseteq X\) is nonempty and bounded below, then if \(\inf S \in X\text{,}\) then

\begin{equation*} \inf S \in \tau_X(S)\text{.} \end{equation*}

Here are the basic facts about the closure operator \(\tau_X \colon \PP(X) \to \PP(X)\text{.}\) As we shall see, these facts are in many ways even more important than the definition of \(\tau_X\) itself.

Let us prove these in order.

  1. If \(x \in S\text{,}\) then for every \(\varepsilon>0\text{,}\) we certainly have \(d(x,x) = 0 \lt \varepsilon\text{.}\)

  2. For every \(\varepsilon>0\text{,}\) the sentence

    \((\exists s \in \varnothing)(d(x,s)\lt \varepsilon)\) is false.

  3. If \(x\) is close to \(S_1 \cup \dots \cup S_n\text{,}\) then for every natural number

    \(m \in \NN^{\ast}\text{,}\) we may select \(s_m \in S_1 \cup \dots \cup S_n\) such that \(d(x, s_m) \lt 1/m\text{.}\) By the pigeonhole principle, there exists an \(i\) such that the set \(\{ m \in \NN^{\ast} : s_m \in S_i \}\) is infinite. Consequently, for any \(\varepsilon>0\text{,}\) there exists a natural number \(m\geq 1\) such that \(s_m \in S_i\) and \(1/m \lt \varepsilon\text{.}\) In particular

    \begin{equation*} d(x,s_m) \lt 1/m \lt \varepsilon\text{.} \end{equation*}

    Thus \(x\) is close to \(S_i\text{.}\) Conversely, if \(x\) is close to some \(S_i\text{,}\) then for every \(\varepsilon >0\text{,}\) there exists \(s \in S_i \subseteq S_1 \cup \dots \cup S_n\) such that \(d(x,s) \lt \varepsilon\text{,}\) so \(x\) is close to \(S_1 \cup \dots \cup S_n\text{.}\)

  4. If \(x\) is close to \(\tau_X(S)\text{,}\) then for any \(\varepsilon >0\text{,}\) we may select \(t\in \tau_X(S)\) such that \(d(x,t) \lt \varepsilon/2\text{.}\) Since \(t\) is close to \(S\text{,}\) we may select \(s \in S\) such that \(d(t, s) \lt \varepsilon/2\text{.}\) Now by the triangle inequality,

    \begin{equation*} d(x, s) \leq d(x, t) + d(t, s) \lt \varepsilon\text{.} \end{equation*}

    Thus \(x\) is close to \(S\text{.}\)

We may write \(T = S \cup (T \smallsetminus S)\text{.}\) Then by the third point of the proposition above,

\begin{equation*} \tau_X(T) = \tau_X(S) \cup \tau_X(T \smallsetminus S)\text{,} \end{equation*}

and so \(\tau_X(S) \subseteq \tau_X(T)\text{.}\)

If \(X \subseteq \RR^n\) is a subspace, then there are two different kinds of subset that play a big role in topology:

Definition 1.1.6.

A subset \(Z \subseteq X\) is closed in \(X\) if and only if

\begin{equation*} \tau_X(Z) = Z\text{.} \end{equation*}

In other words, \(Z\) is closed if and only if every point of \(X\) that is close to \(Z\) is contained in \(Z\text{.}\)

A subset \(U \subseteq X\) is open in \(X\) if and only if \(X \smallsetminus U\) is closed. In other words, \(U\) is open if and only if no point of \(U\) is close to the complement of \(U\text{.}\)

  • For any subspace \(X \subseteq \RR^n\text{,}\) the subset \(X \subseteq X\) is both open and closed.

  • For any subspace \(X \subseteq \RR^n\text{,}\) the subset \(\varnothing \subseteq X\) is both open and closed.

  • In \(\RR\) itself, the subset \([0,1] \subset \RR\) is closed but not open.

  • In \(\RR\text{,}\) the subset \(\left]0,1\right[ \subset \RR\) is open but not closed.

  • In \(\RR\text{,}\) the subsets \(\left[0,1\right[\) and \(\left]0,1\right]\) are neither open nor closed.

  • In the subspace \(\left]0, +\infty\right[ \subseteq \RR\text{,}\) the subset \(\left]0,1\right] \subseteq \left]0, +\infty\right[\) is closed but not open.

  • In the subspace \(\left[0, +\infty\right[ \subseteq \RR\text{,}\) the subset \(\left[0,1\right[ \subseteq \left[0, +\infty\right[\) is open but not closed.

Let \(Z \subseteq X\) be a closed subset that contains \(S\text{.}\) Then

\begin{equation*} \tau_X(S) \subseteq \tau_X(Z) = Z\text{.} \end{equation*}

Hence \(\tau_X(S)\) is contained in any closed subset of \(X\) that contains \(S\text{.}\) It remains to note that \(\tau_X(S)\) is itself a closed subset of \(X\) that contains \(S\text{.}\) By the proposition above,

\begin{equation*} \tau_X(\tau_X(S)) = \tau_X(S)\text{,} \end{equation*}

so \(\tau_X(S)\) is closed.

Suppose that \(U \subseteq X\) is open, and let \(x \in U\) be a point. Then by definition, \(x\) is not close to \(X \smallsetminus U\text{.}\) Hence there exists \(\varepsilon>0\) such that for any point \(y \in X \smallsetminus U\text{,}\) one has \(d(x,y) \geq \varepsilon\text{.}\) That is, \(U\) contains \(B^n(x,\varepsilon) \cap X\text{.}\)

Now assume that for every \(x \in U\text{,}\) there exists an \(\varepsilon >0\) such that

\begin{equation*} B^n(x, \varepsilon) \cap X \subseteq U\text{.} \end{equation*}

We aim to show that \(X \smallsetminus U\) is closed; hence let \(x \in U\) be a point that is close to \(X \smallsetminus U\text{.}\) For some \(\varepsilon>0\text{,}\) the intersection \(B^n(x,\varepsilon) \cap X\) is contained in \(U\text{.}\) But at the same time, since \(x\) is close to \(X \smallsetminus U\text{,}\) the intersection \(B^n(x, \varepsilon) \cap (X \smallsetminus U) \neq \varnothing\text{.}\) This is a contradiction that shows that no point of \(U\) is close to \(X \smallsetminus U\text{;}\) thus \(U\) is open.

Assume that \(Z' \subseteq \RR^n\) is a closed subset. Now consider the intersection \(Z' \cap X\text{.}\) The closure \(\tau_{\RR^n}(Z' \cap X)\) is contained in \(\tau_{\RR^n}(Z') = Z'\text{.}\) Consequently,

\begin{equation*} \tau_X(Z'\cap X) = \tau_{\RR^n}(Z'\cap X) \cap X \subseteq Z' \cap X\text{,} \end{equation*}

so \(Z' \cap X\) is closed.

Conversely, assume that \(Z \subseteq X\) is a closed subset. Then \(Z' \coloneq \tau_{\RR^n}(Z)\) is closed in \(\RR^n\text{,}\) and

\begin{equation*} Z = \tau_X(Z) = \tau_{\RR^n}(Z) \cap X = Z' \cap X\text{.} \end{equation*}

The statements about open subsets of \(X\) now follow by taking complements.

Indeed, if \(U' \subseteq \RR^n\) is an open subset, then let \(Z'\) denote its complement. Since \(Z'\cap X\) is closed by what we've already proved, its complement in \(X\text{,}\) which is \(U' \cap X\) is open. Conversely, if \(U \subseteq X\) is open, then let \(Z\) denote its complement in \(X\text{.}\) By what we've already shown, there is a closed subset \(Z' \subseteq \RR^n\) such that \(Z = Z' \cap X\text{,}\) so if \(U'\) denotes the open complement of \(Z'\text{,}\) then \(U = U'\cap X\text{.}\)

Let \(W = \bigcap_{a\in A} Z_a\text{.}\) We aim to prove that \(\tau_X(W) \subseteq W\text{.}\) Let \(x \in X\) be a point that is close to \(W\text{.}\) For every \(a \in A\text{,}\) since \(W \subseteq Z_a\text{,}\) it follows that \(x\) is close to \(Z_a\) as well. Since \(Z_a\) is closed, \(x \in Z_a\text{.}\) This happens for every \(a \in A\text{,}\) so \(x \in W\text{.}\)

If \(A\) is finite, then we have shown that

\begin{equation*} \tau_X\left(\bigcup_{a \in A} Z_a\right) = \bigcup_{a\in A} \tau_X(Z_a) = \bigcup_{a \in A} Z_a\text{,} \end{equation*}

so the union is closed.

The dual statement follows by forming complements.

Let's consider examples in \(\RR\) in which the finiteness hypotheses of the previous proposition are necessary.

  • Infinite unions of closed subsets need not be closed:

    \begin{equation*} \bigcup_{n\in\NN^{\ast}} \left[-1+\frac{1}{n}, 1-\frac{1}{n}\right] = \left]-1,1\right[\text{,} \end{equation*}

    or open:

    \begin{equation*} \bigcup_{n\in\NN^{\ast}} \left[0,1-\frac{1}{n}\right] = \left[0,1\right[ .\text{.} \end{equation*}

  • Infinite intersections of open subsets need not be open:

    \begin{equation*} \bigcap_{n\in\NN^{\ast}} \left]-1-\frac{1}{n}, 1+\frac{1}{n}\right[ = [-1,1]\text{,} \end{equation*}

    or closed:

    \begin{equation*} \bigcap_{n\in\NN^{\ast}} \left]0, 1+\frac{1}{n}\right[ = \left]0,1\right]\text{.} \end{equation*}

We conclude with some examples of subspaces of Euclidean space to which we will return regularly.

  • For any \(n \in \NN\text{,}\) the \(n\)-sphere \(S^n\) is the subspace

    \begin{equation*} S^n \coloneq \{ x \in \RR^{n+1} : \|x\| = 1 \}\text{.} \end{equation*}

  • The open unit ball is the subspace

    \begin{equation*} B^n \coloneq B^n(0,1) = \{ x \in \RR^n : \| x \| \lt 1 \}\text{,} \end{equation*}

    and its closure in \(\RR^n\) is the closed unit ball

    \begin{equation*} \overline{B}^n \coloneq \{ x \in \RR^n : \| x \| \leq 1 \}\text{.} \end{equation*}

  • If we have a finite collection of subspaces \(X_1 \subseteq \RR^{n_1}, \dots, X_k \subseteq \RR^{n_k}\) then we obtain the subspace

    \begin{equation*} X_1 \times \cdots \times X_k \coloneq \{ x = (x^1, \dots, x^k) \in \RR^{n_1} \times \cdots \times \RR^{n_k} : (\forall i)(x^i \in X_i) \}\text{.} \end{equation*}

  • For any \(n\in \NN\text{,}\) the \(n\)-torus is the subspace

    \begin{equation*} T^n \coloneq S^1 \times \cdots \times S^1 \subseteq \RR^{2n}\text{.} \end{equation*}

    That is,

    \begin{equation*} T^n \coloneq \{ (x_1, y_1, \dots, x_n, y_n) \in \RR^{2n} : (\forall i)(x_i^2 +y_i^2=1) \}\text{.} \end{equation*}