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Section 7.2 Open and closed sets

Note: 2 lectures

Subsection 7.2.1 Topology

Before we get to convergence, we define the so-called topology. That is, we define open and closed sets in a metric space. And before doing that, we define two special open and closed sets.

Definition 7.2.1.

Let \((X,d)\) be a metric space, \(x \in X\text{,}\) and \(\delta > 0\text{.}\) Define the open ball, or simply ball, of radius \(\delta\) around \(x\) as

\begin{equation*} B(x,\delta) := \bigl\{ y \in X : d(x,y) < \delta \bigr\} . \end{equation*}

Define the closed ball as

\begin{equation*} C(x,\delta) := \bigl\{ y \in X : d(x,y) \leq \delta \bigr\} . \end{equation*}

When dealing with different metric spaces, it is sometimes vital to emphasize which metric space the ball is in. We do this by writing \(B_X(x,\delta) := B(x,\delta)\) or \(C_X(x,\delta) := C(x,\delta)\text{.}\)

Example 7.2.2.

Take the metric space \(\R\) with the standard metric. For \(x \in \R\) and \(\delta > 0\text{,}\)

\begin{equation*} B(x,\delta) = (x-\delta,x+\delta) \qquad \text{and} \qquad C(x,\delta) = [x-\delta,x+\delta] . \end{equation*}

Example 7.2.3.

Be careful when working on a subspace. Consider the metric space \([0,1]\) as a subspace of \(\R\text{.}\) Then in \([0,1]\text{,}\)

\begin{equation*} B(0,\nicefrac{1}{2}) = B_{[0,1]}(0,\nicefrac{1}{2}) = \bigl\{ y \in [0,1] : \abs{0-y} < \nicefrac{1}{2} \bigr\} = [0,\nicefrac{1}{2}) . \end{equation*}

This is different from \(B_{\R}(0,\nicefrac{1}{2}) = (\nicefrac{-1}{2},\nicefrac{1}{2})\text{.}\) The important thing to keep in mind is which metric space we are working in.

Definition 7.2.4.

Let \((X,d)\) be a metric space. A subset \(V \subset X\) is open if for every \(x \in V\text{,}\) there exists a \(\delta > 0\) such that \(B(x,\delta) \subset V\text{.}\) See Figure 7.4. A subset \(E \subset X\) is closed if the complement \(E^c = X \setminus E\) is open. When the ambient space \(X\) is not clear from context, we say \(V\) is open in \(X\) and \(E\) is closed in \(X\).

If \(x \in V\) and \(V\) is open, then we say \(V\) is an open neighborhood of \(x\) (or sometimes just neighborhood).


Figure 7.4. Open set in a metric space. Note that \(\delta\) depends on \(x\text{.}\)

Intuitively, an open set \(V\) is a set that does not include its “boundary.” Wherever we are in \(V\text{,}\) we are allowed to “wiggle” a little bit and stay in \(V\text{.}\) Similarly, a set \(E\) is closed if everything not in \(E\) is some distance away from \(E\text{.}\) The open and closed balls are examples of open and closed sets (this must still be proved). But not every set is either open or closed. Generally, most subsets are neither.

Example 7.2.5.

The set \((0,\infty) \subset \R\) is open: Given any \(x \in (0,\infty)\text{,}\) let \(\delta := x\text{.}\) Then \(B(x,\delta) = (0,2x) \subset (0,\infty)\text{.}\)

The set \([0,\infty) \subset \R\) is closed: Given \(x \in (-\infty,0) = [0,\infty)^c\text{,}\) let \(\delta := -x\text{.}\) Then \(B(x,\delta) = (-2x,0) \subset (-\infty,0) = [0,\infty)^c\text{.}\)

The set \([0,1) \subset \R\) is neither open nor closed. First, every ball in \(\R\) around \(0\text{,}\) \(B(0,\delta) = (-\delta,\delta)\text{,}\) contains negative numbers and hence is not contained in \([0,1)\text{.}\) So \([0,1)\) is not open. Second, every ball in \(\R\) around \(1\text{,}\) \(B(1,\delta) = (1-\delta,1+\delta)\text{,}\) contains numbers strictly less than 1 and greater than 0 (e.g. \(1-\nicefrac{\delta}{2}\) as long as \(\delta < 2\)). Thus \([0,1)^c = \R \setminus [0,1)\) is not open, and \([0,1)\) is not closed.

The index set \(I\) in iii can be arbitrarily large. By \(\bigcup_{\lambda \in I} V_\lambda\text{,}\) we simply mean the set of all \(x\) such that \(x \in V_\lambda\) for at least one \(\lambda \in I\text{.}\)

Proof.

The sets \(\emptyset\) and \(X\) are obviously open in \(X\text{.}\)

Let us prove ii. If \(x \in \bigcap_{j=1}^k V_j\text{,}\) then \(x \in V_j\) for all \(j\text{.}\) As \(V_j\) are all open, for every \(j\) there exists a \(\delta_j > 0\) such that \(B(x,\delta_j) \subset V_j\text{.}\) Take \(\delta := \min \{ \delta_1,\delta_2,\ldots,\delta_k \}\) and notice \(\delta > 0\text{.}\) We have \(B(x,\delta) \subset B(x,\delta_j) \subset V_j\) for every \(j\) and so \(B(x,\delta) \subset \bigcap_{j=1}^k V_j\text{.}\) Consequently the intersection is open.

Let us prove iii. If \(x \in \bigcup_{\lambda \in I} V_\lambda\text{,}\) then \(x \in V_\lambda\) for some \(\lambda \in I\text{.}\) As \(V_\lambda\) is open, there exists a \(\delta > 0\) such that \(B(x,\delta) \subset V_\lambda\text{.}\) But then \(B(x,\delta) \subset \bigcup_{\lambda \in I} V_\lambda\text{,}\) and so the union is open.

Example 7.2.7.

The main thing to notice is the difference between items ii and iii. Item ii is not true for an arbitrary intersection. For example, \(\bigcap_{n=1}^\infty (\nicefrac{-1}{n},\nicefrac{1}{n}) = \{ 0 \}\text{,}\) which is not open.

The proof of the following analogous proposition for closed sets is left as an exercise.

Despite the naming, we have not yet shown that the open ball is open and the closed ball is closed. Let us show these facts now to justify the terminology.

Proof.

Let \(y \in B(x,\delta)\text{.}\) Let \(\alpha := \delta-d(x,y)\text{.}\) As \(\alpha > 0\text{,}\) consider \(z \in B(y,\alpha)\text{.}\) Then

\begin{equation*} d(x,z) \leq d(x,y) + d(y,z) < d(x,y) + \alpha = d(x,y) + \delta-d(x,y) = \delta . \end{equation*}

Therefore, \(z \in B(x,\delta)\) for every \(z \in B(y,\alpha)\text{.}\) So \(B(y,\alpha) \subset B(x,\delta)\) and \(B(x,\delta)\) is open. See Figure 7.5.


Figure 7.5. Proof that \(B(x,\delta)\) is open: \(B(y,\alpha) \subset B(x,\delta)\) with the triangle inequality illustrated.

The proof that \(C(x,\delta)\) is closed is left as an exercise.

Again, be careful about which metric space we are in. The set \([0,\nicefrac{1}{2})\) is an open ball in \([0,1]\text{,}\) and so \([0,\nicefrac{1}{2})\) is an open set in \([0,1]\text{.}\) On the other hand, \([0,\nicefrac{1}{2})\) is neither open nor closed in \(\R\text{.}\)

The proof is left as an exercise. Keep in mind that there are many other open and closed sets in the set of real numbers.

For example, let \(X := \R\text{,}\) \(Y:=[0,1]\text{,}\) \(U := [0,\nicefrac{1}{2})\text{.}\) We saw that \(U\) is an open set in \(Y\text{.}\) We may take \(V := (\nicefrac{-1}{2},\nicefrac{1}{2})\text{.}\)

Proof.

Suppose \(V \subset X\) is open and \(x \in V \cap Y\text{.}\) Let \(U := V \cap Y\text{.}\) As \(V\) is open, there exists a \(\delta > 0\) such that \(B_X(x,\delta) \subset V\text{.}\) Then

\begin{equation*} B_Y(x,\delta) = B_X(x,\delta) \cap Y \subset V \cap Y = U . \end{equation*}

The proof of the opposite direction, that is, that if \(U \subset Y\) is open in the subspace topology there exists a \(V\) is left as Exercise 7.2.12.

A hint for finshing the proof (the exercise) is that a useful way to think about an open set is as a union of open balls. If \(U\) is open, then for each \(x \in U\text{,}\) there is a \(\delta_x > 0\) (depending on \(x\)) such that \(B(x,\delta_x) \subset U\text{.}\) Then \(U = \bigcup_{x\in U} B(x,\delta_x)\text{.}\)

In case of an open subset of an open set or a closed subset of a closed set, matters are simpler.

Proof.

Let us prove i and leave ii to an exercise.

If \(U \subset V\) is open in the subspace topology, by Proposition 7.2.11, there exists a set \(W \subset X\) open in \(X\text{,}\) such that \(U = W \cap V\text{.}\) Intersection of two open sets is open so \(U\) is open in \(X\text{.}\)

Now suppose \(U\) is open in \(X\text{.}\) Then \(U = U \cap V\text{.}\) So \(U\) is open in \(V\) again by Proposition 7.2.11.

Subsection 7.2.2 Connected sets

Let us generalize the idea of an interval to general metric spaces. One of the main features of an interval in \(\R\) is that it is connected—that we can continuously move from one point of it to another point without jumping. For example, in \(\R\) we usually study functions on intervals, and in more general metric spaces we usually study functions on connected sets.

Definition 7.2.13.

A nonempty 1  metric space \((X,d)\) is connected if the only subsets of \(X\) that are both open and closed (so-called clopen subsets) are \(\emptyset\) and \(X\) itself. If a nonempty \((X,d)\) is not connected we say it is disconnected.

When we apply the term connected to a nonempty subset \(A \subset X\text{,}\) we mean that \(A\) with the subspace topology is connected.

In other words, a nonempty \(X\) is connected if whenever we write \(X = X_1 \cup X_2\) where \(X_1 \cap X_2 = \emptyset\) and \(X_1\) and \(X_2\) are open, then either \(X_1 = \emptyset\) or \(X_2 = \emptyset\text{.}\) So to show \(X\) is disconnected, we need to find nonempty disjoint open sets \(X_1\) and \(X_2\) whose union is \(X\text{.}\) For subsets, we state this idea as a proposition. The proposition is illustrated in Figure 7.6.


Figure 7.6. Disconnected subset. Notice that \(U_1 \cap U_2\) need not be empty, but \(U_1 \cap U_2 \cap S = \emptyset\text{.}\)

Proof.

First suppose \(S\) is disconnected: There are nonempty disjoint \(S_1\) and \(S_2\) that are open in \(S\) and \(S = S_1 \cup S_2\text{.}\) Proposition 7.2.11 says there exist \(U_1\) and \(U_2\) that are open in \(X\) such that \(U_1 \cap S = S_1\) and \(U_2 \cap S = S_2\text{.}\)

For the other direction start with the \(U_1\) and \(U_2\text{.}\) Then \(U_1 \cap S\) and \(U_2 \cap S\) are open in \(S\) by Proposition 7.2.11. Via the discussion before the proposition, \(S\) is disconnected.

Example 7.2.15.

Let \(S \subset \R\) be such that \(x < z < y\) with \(x,y \in S\) and \(z \notin S\text{.}\) Claim: \(S\) is disconnected. Proof: Notice

\begin{equation*} \bigl( (-\infty,z) \cap S \bigr) \cup \bigl( (z,\infty) \cap S \bigr) = S . \end{equation*}

Proof.

Suppose \(S\) is connected. If \(S\) is a single point, then we are done. So suppose \(x < y\) and \(x,y \in S\text{.}\) If \(z \in \R\) is such that \(x < z < y\text{,}\) then \((-\infty,z) \cap S\) is nonempty and \((z,\infty) \cap S\) is nonempty. The two sets are disjoint. As \(S\) is connected, we must have they their union is not \(S\text{,}\) so \(z \in S\text{.}\) By Proposition 1.4.1, \(S\) is an interval.

If \(S\) is a single point, it is connected. Therefore, suppose \(S\) is an interval. Consider open subsets \(U_1\) and \(U_2\) of \(\R\text{,}\) such that \(U_1 \cap S\) and \(U_2 \cap S\) are nonempty, and \(S = \bigl( U_1 \cap S \bigr) \cup \bigl( U_2 \cap S \bigr)\text{.}\) We will show that \(U_1 \cap S\) and \(U_2 \cap S\) contain a common point, so they are not disjoint, proving that \(S\) is connected. Suppose \(x \in U_1 \cap S\) and \(y \in U_2 \cap S\text{.}\) Without loss of generality, assume \(x < y\text{.}\) As \(S\) is an interval, \([x,y] \subset S\text{.}\) Note that \(U_2 \cap [x,y] \not= \emptyset\text{,}\) and let \(z := \inf (U_2 \cap [x,y])\text{.}\) We wish to show that \(z \in U_1\text{.}\) If \(z = x\text{,}\) then \(z \in U_1\text{.}\) If \(z > x\text{,}\) then for every \(\epsilon > 0\text{,}\) the ball \(B(z,\epsilon) = (z-\epsilon,z+\epsilon)\) contains points of \([x,y]\) not in \(U_2\text{,}\) as \(z\) is the infimum of such points. So \(z \notin U_2\) as \(U_2\) is open. Therefore, \(z \in U_1\) as every point of \([x,y]\) is in \(U_1\) or \(U_2\text{.}\) As \(U_1\) is open, \(B(z,\delta) \subset U_1\) for a small enough \(\delta > 0\text{.}\) As \(z\) is the infimum of the nonempty set \(U_2 \cap [x,y]\text{,}\) there must exist some \(w \in U_2 \cap [x,y]\) such that \(w \in [z,z+\delta) \subset B(z,\delta) \subset U_1\text{.}\) Therefore, \(w \in U_1 \cap U_2 \cap [x,y]\text{.}\) So \(U_1 \cap S\) and \(U_2 \cap S\) are not disjoint, and \(S\) is connected. See Figure 7.7.


Figure 7.7. Proof that an interval is connected.

Example 7.2.17.

Oftentimes a ball \(B(x,\delta)\) is connected, but this is not necessarily true in every metric space. For a simplest example, take a two point space \(\{ a, b\}\) with the discrete metric. Then \(B(a,2) = \{ a , b \}\text{,}\) which is not connected as \(B(a,1) = \{ a \}\) and \(B(b,1) = \{ b \}\) are open and disjoint.

Subsection 7.2.3 Closure and boundary

Sometimes we wish to take a set and throw in everything that we can approach from the set. This concept is called the closure.

Definition 7.2.18.

Let \((X,d)\) be a metric space and \(A \subset X\text{.}\) The closure of \(A\) is the set

\begin{equation*} \widebar{A} := \bigcap \{ E \subset X : E \text{ is closed and } A \subset E \} . \end{equation*}

That is, \(\widebar{A}\) is the intersection of all closed sets that contain \(A\text{.}\)

Proof.

The closure is an intersection of closed sets, so \(\widebar{A}\) is closed. There is at least one closed set containing \(A\text{,}\) namely \(X\) itself, so \(A \subset \widebar{A}\text{.}\) If \(A\) is closed, then \(A\) is a closed set that contains \(A\text{.}\) So \(\widebar{A} \subset A\text{,}\) and thus \(A = \widebar{A}\text{.}\)

Example 7.2.20.

The closure of \((0,1)\) in \(\R\) is \([0,1]\text{.}\) Proof: If \(E\) is closed and contains \((0,1)\text{,}\) then \(E\) contains \(0\) and \(1\) (why?). Thus \([0,1] \subset E\text{.}\) But \([0,1]\) is also closed. Hence, the closure \(\overline{(0,1)} = [0,1]\text{.}\)

Example 7.2.21.

Be careful to notice what ambient metric space you are working with. If \(X = (0,\infty)\text{,}\) then the closure of \((0,1)\) in \((0,\infty)\) is \((0,1]\text{.}\) Proof: Similarly as above, \((0,1]\) is closed in \((0,\infty)\) (why?). Any closed set \(E\) that contains \((0,1)\) must contain 1 (why?). Therefore, \((0,1] \subset E\text{,}\) and hence \(\overline{(0,1)} = (0,1]\) when working in \((0,\infty)\text{.}\)

Let us justify the statement that the closure is everything that we can “approach” from the set.

Proof.

Let us prove the two contrapositives. Let us show that \(x \notin \widebar{A}\) if and only if there exists a \(\delta > 0\) such that \(B(x,\delta) \cap A = \emptyset\text{.}\)

First suppose \(x \notin \widebar{A}\text{.}\) We know \(\widebar{A}\) is closed. Thus there is a \(\delta > 0\) such that \(B(x,\delta) \subset \widebar{A}^c\text{.}\) As \(A \subset \widebar{A}\) we see that \(B(x,\delta) \subset \widebar{A}^c \subset A^c\) and hence \(B(x,\delta) \cap A = \emptyset\text{.}\)

On the other hand, suppose there is a \(\delta > 0\text{,}\) such that \(B(x,\delta) \cap A = \emptyset\text{.}\) In other words, \(A \subset {B(x,\delta)}^c\text{.}\) As \({B(x,\delta)}^c\) is a closed set, as \(x \not \in {B(x,\delta)}^c\text{,}\) and as \(\widebar{A}\) is the intersection of closed sets containing \(A\text{,}\) we have \(x \notin \widebar{A}\text{.}\)

We can also talk about the interior of a set (points we cannot approach from the complement), and the boundary of a set (points we can approach both from the set and its complement).

Definition 7.2.23.

Let \((X,d)\) be a metric space and \(A \subset X\text{.}\) The interior of \(A\) is the set

\begin{equation*} A^\circ := \{ x \in A : \text{there exists a } \delta > 0 \text{ such that } B(x,\delta) \subset A \} . \end{equation*}

The boundary of \(A\) is the set

\begin{equation*} \partial A := \widebar{A}\setminus A^\circ. \end{equation*}

Alternatively, the interior is the union of open sets lying in \(A\text{,}\) see Exercise 7.2.14. By definition, \(A^\circ \subset A\text{;}\) however, the points of the boundary may or may not be in \(A\text{.}\)

Example 7.2.24.

Suppose \(A:=(0,1]\) and \(X := \R\text{.}\) Then \(\widebar{A}=[0,1]\text{,}\) \(A^\circ = (0,1)\text{,}\) and \(\partial A = \{ 0, 1 \}\text{.}\)

Example 7.2.25.

Consider \(X := \{ a, b \}\) with the discrete metric, and let \(A := \{ a \}\text{.}\) Then \(\widebar{A} = A^\circ = A\) and \(\partial A = \emptyset\text{.}\)

Proof.

Given \(x \in A^\circ\text{,}\) there is a \(\delta > 0\) such that \(B(x,\delta) \subset A\text{.}\) If \(z \in B(x,\delta)\text{,}\) then as open balls are open, there is an \(\epsilon > 0\) such that \(B(z,\epsilon) \subset B(x,\delta) \subset A\text{.}\) So \(z \in A^\circ\text{.}\) Therefore, \(B(x,\delta) \subset A^\circ\text{,}\) and \(A^\circ\) is open.

As \(A^\circ\) is open, then \(\partial A = \widebar{A} \setminus A^\circ = \widebar{A} \cap {(A^\circ)}^c\) is closed.

The boundary is the set of points that are close to both the set and its complement. See Figure 7.8 for the a diagram of the next proposition.


Figure 7.8. Boundary is the set where every ball contains points in the set and also its complement.

Proof.

Suppose \(x \in \partial A = \widebar{A} \setminus A^\circ\) and let \(\delta > 0\) be arbitrary. By Proposition 7.2.22, \(B(x,\delta)\) contains a point of \(A\text{.}\) If \(B(x,\delta)\) contained no points of \(A^c\text{,}\) then \(x\) would be in \(A^\circ\text{.}\) Hence \(B(x,\delta)\) contains a point of \(A^c\) as well.

Let us prove the other direction by contrapositive. Suppose \(x \notin \partial A\text{,}\) so \(x \notin \widebar{A}\) or \(x \in A^\circ\text{.}\) If \(x \notin \widebar{A}\text{,}\) then \(B(x,\delta) \subset \widebar{A}^c\) for some \(\delta > 0\) as \(\widebar{A}\) is closed. So \(B(x,\delta) \cap A\) is empty, because \(\widebar{A}^c \subset A^c\text{.}\) If \(x \in A^\circ\text{,}\) then \(B(x,\delta) \subset A\) for some \(\delta > 0\text{,}\) so \(B(x,\delta) \cap A^c\) is empty.

We obtain the following immediate corollary about closures of \(A\) and \(A^c\text{.}\) We simply apply Proposition 7.2.22.

Subsection 7.2.4 Exercises

Exercise 7.2.2.

Finish the proof of Proposition 7.2.9 by proving that \(C(x,\delta)\) is closed.

Exercise 7.2.4.

Suppose \((X,d)\) is a nonempty metric space with the discrete topology. Show that \(X\) is connected if and only if it contains exactly one element.

Exercise 7.2.5.

Take \(\Q\) with the standard metric, \(d(x,y) = \abs{x-y}\text{,}\) as our metric space. Prove that \(\Q\) is totally disconnected, that is, show that for every \(x, y \in \Q\) with \(x \not= y\text{,}\) there exists an two open sets \(U\) and \(V\text{,}\) such that \(x \in U\text{,}\) \(y \in V\text{,}\) \(U \cap V = \emptyset\text{,}\) and \(U \cup V = \Q\text{.}\)

Exercise 7.2.6.

Show that in a metric space, every open set can be written as a union of closed sets.

Exercise 7.2.7.

Prove that in a metric space,

  1. \(E\) is closed if and only if \(\partial E \subset E\text{.}\)

  2. \(U\) is open if and only if \(\partial U \cap U = \emptyset\text{.}\)

Exercise 7.2.8.

Prove that in a metric space,

  1. \(A\) is open if and only if \(A^\circ = A\text{.}\)

  2. \(U \subset A^\circ\) for every open set \(U\) such that \(U \subset A\text{.}\)

Exercise 7.2.9.

Let \(X\) be a set and \(d\text{,}\) \(d'\) be two metrics on \(X\text{.}\) Suppose there exists an \(\alpha > 0\) and \(\beta > 0\) such that \(\alpha d(x,y) \leq d'(x,y) \leq \beta d(x,y)\) for all \(x,y \in X\text{.}\) Show that \(U\) is open in \((X,d)\) if and only if \(U\) is open in \((X,d')\text{.}\) That is, the topologies of \((X,d)\) and \((X,d')\) are the same.

Exercise 7.2.10.

Suppose \(\{ S_i \}\text{,}\) \(i \in \N\text{,}\) is a collection of connected subsets of a metric space \((X,d)\text{,}\) and there exists an \(x \in X\) such that \(x \in S_i\) for all \(i \in \N\text{.}\) Show that \(\bigcup_{i=1}^\infty S_i\) is connected.

Exercise 7.2.11.

Let \(A\) be a connected set in a metric space.

  1. Is \(\widebar{A}\) connected? Prove or find a counterexample.

  2. Is \(A^\circ\) connected? Prove or find a counterexample.

Hint: Think of sets in \(\R^2\text{.}\)

Exercise 7.2.12.

Finish the proof of Proposition 7.2.11. Suppose \((X,d)\) is a metric space and \(Y \subset X\text{.}\) Show that with the subspace metric on \(Y\text{,}\) if a set \(U \subset Y\) is open (in \(Y\)), then there exists an open set \(V \subset X\) such that \(U = V \cap Y\text{.}\)

Exercise 7.2.13.

Let \((X,d)\) be a metric space.

  1. For every \(x \in X\) and \(\delta > 0\text{,}\) show \(\overline{B(x,\delta)} \subset C(x,\delta)\text{.}\)

  2. Is it always true that \(\overline{B(x,\delta)} = C(x,\delta)\text{?}\) Prove or find a counterexample.

Exercise 7.2.14.

Let \((X,d)\) be a metric space and \(A \subset X\text{.}\) Show that \(A^\circ = \bigcup \{ V : V \text{ is open and } V \subset A \}\text{.}\)

Exercise 7.2.16.

Let \((X,d)\) be a metric space. Show that there exists a bounded metric \(d'\) such that \((X,d')\) has the same open sets, that is, the topology is the same.

Exercise 7.2.17.

Let \((X,d)\) be a metric space.

  1. Prove that for every \(x \in X\text{,}\) there either exists a \(\delta > 0\) such that \(B(x,\delta) = \{ x \}\text{,}\) or \(B(x,\delta)\) is infinite for every \(\delta > 0\text{.}\)

  2. Find an explicit example of \((X,d)\text{,}\) \(X\) infinite, where for every \(\delta > 0\) and every \(x \in X\text{,}\) the ball \(B(x,\delta)\) is finite.

  3. Find an explicit example of \((X,d)\) where for every \(\delta > 0\) and every \(x \in X\text{,}\) the ball \(B(x,\delta)\) is countably infinite.

  4. Prove that if \(X\) is uncountable, then there exists an \(x \in X\) and a \(\delta > 0\) such that \(B(x,\delta)\) is uncountable.

Exercise 7.2.18.

For every \(x \in \R^n\) and every \(\delta > 0\) define the “rectangle” \(R(x,\delta) := (x_1-\delta,x_1+\delta) \times (x_2-\delta,x_2+\delta) \times \cdots \times (x_n-\delta,x_n+\delta)\text{.}\) Show that these sets generate the same open sets as the balls in standard metric. That is, show that a set \(U \subset \R^n\) is open in the sense of the standard metric if and only if for every point \(x \in U\text{,}\) there exists a \(\delta > 0\) such that \(R(x,\delta) \subset U\text{.}\)

Some authors do not exclude the empty set from the definition, and the empty set would then be connected. We avoid the empty set for essentially the same reason why 1 is neither a prime nor a composite number: Our connected sets have exactly two clopen subsets and disconnected sets have more than two. The empty set has exactly one.
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