Compactness#
The final part of the course is about compactness. Compactness is a slightly less intuitive property than, for example, connectivity. Roughly speaking, compactness can be viewed as a generalisation of what it means for a subset of \(\mathbb{R}^n\) to be closed and bounded, to other topological spaces.
Compact spaces#
The notion of compactness relies on open covers as defined below.
Definition 106 (Open cover)
Let \(\mathcal{U} \subseteq 2^X\) be a topology on \(X.\) An open cover of \(X\) is a subset \(\mathcal{V} \subseteq \mathcal{U}\) such that
We say \(\mathcal{V}\) covers \(X.\) If \(\mathcal{V}' \subseteq \mathcal{V}\) and \(\mathcal{V}'\) covers \(X,\) then we say \(\mathcal{V}'\) is a subcover of \(\mathcal{V}.\)
Now, a space is defined to be compact if any open cover of the space contains a finite subcover.
Definition 107 (Compact space)
A topological space \(X\) is compact if every open cover \(\mathcal{V}\) of \(X\) has a finite subcover \(\mathcal{V}' = \{V_1, \dots, V_n\} \subseteq \mathcal{V}.\)
We note that compactness is topological property: if a compact space is homeomorphic to another space, the second space is also compact. The first result that we show is that the closed unit interval is compact.
Theorem 98 (Closed interval is compact)
The closed interval \([0, 1] \subseteq \mathbb{R}\) with the standard topology is compact.
Closed interval is compact
Suppose \(\mathcal{V}\) is an open cover of \([0, 1].\) Define
We first show that \(A\) is non-empty. Since \(\mathcal{V}\) is an open cover of \([0, 1],\) there exists \(V \in \mathcal{V}\) that contains \(0.\) Therefore \(\{0\}\) has a finite sub-cover from \(\mathcal{V}\) and \(0 \in A.\)
Now, let \(\alpha = \sup A.\) Suppose \(\alpha < 1.\) Since \(\mathcal{V}\) covers \([0, 1],\) there exists open \(V_\alpha \in \mathcal{V}\) with \(\alpha \in V_\alpha.\) Since \(V_\alpha\) is open, there exists \(\epsilon > 0\) such that \(B_\epsilon(\alpha) \subseteq V_\alpha.\) By the definition of \(\alpha,\) the set \([0, \max(0, \alpha - \epsilon / 2)]\) has a finite subcover, to which we can add \(V_\alpha\) to get a finite subcover of \([0, \min(1, \alpha + \epsilon / 2)].\) This leads to a contradiction, so \(\alpha = 1.\)
We finally show that \(\alpha \in A.\) Repeating this argument, since \(\mathcal{V}\) covers \([0, 1],\) there exists open \(V_1 \in \mathcal{V}\) that contains \(1,\) such that \((1 - \epsilon, 1] \subseteq V_1.\) Since \(1 - \epsilon / 2 \in A,\) there exists a finite \(\mathcal{V}' \subseteq \mathcal{V}\) which covers \([0, 1 - \epsilon].\) Then \(\mathcal{W} = \mathcal{V}' \cup \{V_1\}\) is a finite subcover of \(\mathcal{V}.\)
Note how this proof relies on the fact that the open interval is closed and bounded. For example, \(\mathbb{R}\) is not compact because the set \(\{(n, n+2): n \in \mathbb{N}\}\) is an open cover that does not contain a finite subcover. Similarly, the open interval \((0, 1)\) is not compact even though it is bounded, since the set \(\{(n^{-1}, 1): n \in \mathbb{N}\}\) is an open cover that does not contain a finite subcover.
We now show two useful results. First, we note that a subset of a compact space is not necessarily a compact space: \((0, 1) \subseteq [0, 1]\) but the former is not compact even though the latter is. However, a closed subset of a compact space is compact.
Lemma 35 (Closed subsets of compact spaces are compact)
If \(X\) is compact and \(C\) is a closed subset of \(X,\) then \(C\) is also compact.
Proof: Closed subsets of compact spaces are compact
Suppose \(X\) is compact and \(C\) is a closed subset of \(X.\) Let \(\mathcal{V}\) be an open cover of \(C.\) Then since each \(V \in \mathcal{V}\) is open in \(C,\) there exists open \(V'\) in \(X\) such that \(V' \cap C = V.\) Then, the set \(\mathcal{W}\) containing all such \(V',\) i.e. all open \(V'\) such that \(V' \cap C = V\) for some \(V \in \mathcal{V},\) together with the open set \(X \setminus C,\) is an open cover of \(X.\) Since \(X\) is compact, it has a finite subcover from \(\mathcal{W},\) and since \(C \subseteq X,\) this also gives a finite subcover for \(C,\) which is therefore compact.
Second, a similar result holds for closedness: a compact subspace of a Hausdorff space is closed.
Lemma 36 (Compact subspaces of Hausdorff spaces are closed)
Let \(X\) be a Hausdorff space. If \(C \subseteq X\) is compact, then \(C\) is closed in \(X.\)
Proof: Compact subspaces of Hausdorff spaces are closed
Let \(X\) be a Hausdorff space. Suppose \(C \subseteq X\) is compact. We will show that \(X \setminus C\) is open.
Suppose \(x \in X \setminus C.\) We will show there exists an open \(U \subseteq X \setminus C\) that contains \(x.\) Since \(X\) is Hausdorff, for each \(y \in C\) there exists an open \(W_{xy}\) containing \(y\) and an open \(U_{xy}\) containing \(x\) such that \(W_{xy} \cap U_{xy} = \emptyset.\) Then, the collection \(\{W_{xy} \subseteq X: y \in C\}\) is an open cover of \(C,\) and since \(C\) is compact, it has a finite subcover of the form \(\{W_{x_1y}, \dots, W_{x_Ny}\}.\) Then the set \(\cap_{n = 1}^N U_{x_ny}\) is open because it is a finite interesection of open sets, and it does not intersect \(C,\) so it is contained in \(X \setminus C.\) Therefore \(X \setminus C\) is open and \(C\) is closed.
It turns out that for \(\mathbb{R}^n\) with the standard topology, compactness is equivalent to closedness plus boundedness. To show this, we first define boundedness for metric spaces.
Definition 108 (Bounded metric space)
A metric space \((X, d)\) is bounded if there exists \(M \in \mathbb{R}\) such that \(d(x, y) \leq M\) for all \(x, y \in X.\)
Note that being bounded is not a topological property. For example \((0, 1) \simeq \mathbb{R}\) are homeomorphic but \((0, 1)\) is bounded while \(\mathbb{R}\) is not. A useful intermediate result is that a closed metric space is bounded.
Lemma 37 (Compact metric spaces are bounded)
A compact metric space \((X, d)\) is bounded.
Proof: Compact metric spaces are bounded
Suppose \((X, d)\) is compact. Pick \(x \in X.\) Then \(V = \{B_r(x) : r \in \mathbb{R}^+\}\) is an open cover of \(X.\) Since \(X\) is compact, there exists a finite subcover of the form \(\{B_{r_1}(x), \dots, B_{r_n}(x)\}.\) Let \(R = \max\{r_1, \dots, r_n\}.\) Then \(d(x, y) < R\) for all \(y \in X,\) so for all \(y, z \in X\) we have
Therefore \(X\) is boudnded.
We now arrive at one of the main results of this section, the Heine Borel theorem, which states a subsets of \(\mathbb{R}\) is compact if and only if it is closed and bounded.
Theorem 99 (Heine-Borel)
A subset \(C \subseteq \mathbb{R}\) with the standard topology is compact if and only if it is closed and bounded.
Proof: Heine-Borel
Suppose \(C \subseteq \mathbb{R}\) with the standard topology.
Implies: Suppose \(C\) is closed and bounded. Since \(C\) is bounded, it is contained in a closed set \([-M, M]\) for some \(M \in \mathbb{R}.\) Therefore \(C\) is a closed subset of a compact space \([-M, M] \simeq [0, 1],\) therefore it is compact.
Implied by: Suppose \(C\) is compact. Since \(\mathbb{R}\) is a metric space, it is also Hausdorff, and since \(C\) is compact it follows that it is closed (Lemma 35). In addition, since \(C\) is a compact metric space, it is also bounded ().
The Heine-Borel theorem formalises the earlier point we made about compactness being a generalisation of closedness plus boundedness, beyond the reals.
Lemma 38 (Compact subsets of \(\mathbb{R}\) have finite upper bounds)
If \(A \subseteq \mathbb{R}\) with the standard topology is compact, there exists \(\alpha \in A\) such that \(\alpha \geq a\) for all \(a \in A.\)
Proof: Compact subsets of \(~\mathbb{R}~\) have finite upper bounds
Suppose \(A \subseteq \mathbb{R}\) with the standard topology is compact. Since \(A\) is compact, it is bounded and closed. Let \(\alpha = \sup A.\) Then \(\alpha \geq a\) for all \(a \in A,\) and it remains to show that \(\alpha \in A.\)
Suppose that \(\alpha \not \in A.\) Then \(x \in \mathbb{R} \setminus A\) and since \(A\) is closed, \(\mathbb{R} \setminus A\) is open. By the definition of supremum, for each \(\epsilon > 0,\) there exists \(a \in A\) such that \(|\alpha - a| < \epsilon.\) This is a contradiction, because there must exist a sufficiently small ball centered on \(\alpha\) that is contained in \(\mathbb{R} \setminus A,\) because the latter is open. Therefore \(\alpha \in A.\)
Lemma 39 (Image of compact space under continuous function is compact)
Let \(X\) and \(Y\) be topological spaces. If \(f: X \to Y\) is continuous and \(X\) is compact, then \(\text{im}f \subseteq\) is also compact.
Proof: Image of compact space under continuous function is compact
Let \(X\) and \(Y\) be topological spaces. Suppose \(f: X \to Y\) is continuous and \(X\) is compact.
Let \(\mathcal{V} = \{V_\alpha \subseteq Y: \alpha \in T\}\) be an open cover of \(\text{im}f.\) Then the sets \(f^{-1}(V_\alpha), \alpha \in T\) are open in \(X\) and also form an open cover of it. Since \(X\) is compact, it has a finite subcover of the form \(\{f^{-1}(V_{\alpha_1}), \dots, f^{-1}(V_{\alpha_n})\}.\) We conclude that \(V_{\alpha_1}, \dots, V_{\alpha_N}\) form a finite subcover of \(\text{im}f,\) which is therefore comapct.
Definition 109 (Maximum value theorem)
If \(X\) is a compact topological space and \(f: X \to \mathbb{R}\) is continuous with the standard topology, then \(x \in X\) such that \(f(x) \geq f(x')\) for all \(x' \in X.\)
Proof: Maximum value theorem
Suppose \(X\) is a topological space and let \(f: X \to \mathbb{R}\) be continuous. The image of a compact set under a continuous function is compact (Lemma 39), so \(\text{im}f\) is compact. Since \(\mathbb{R}\) is a metric space, so \(\text{im}f\) is also a metric space and is therefore Hausdorff. Because \(\text{im}f\) is a compact metric space, it must be bounded (Lemma 37). Therefore \(\text{im} f\) contains its supremum, i.e. \(f(x) = \sup \text{im} f\) for some \(x \in X.\) By definition of the supremum, \(f(x) \geq f(x')\) for all \(x' \in X.\)
Lemma 40 (Maximum value theorem for \([0, 1]\))
If \(f: [0, 1] \to \mathbb{R}\) is continuous with the standard topology on both \([0, 1]\) and \(\mathbb{R},\) then there exists \(x \in X\) such that \(f(x) \geq f(x')\) for all \(x' \in X.\)
Maximum value theorem for \(~[0, 1]\)
This follows from Definition 109, since the closed interval \([0, 1]\) with the standard topology is compact.
Products and quotients#
Now we turn to some properties of products and quotients of compact spaces. The results in this section are very useful for proving many other properties.
Product spaces#
One particularly useful result is that the product of compact spaces is compact.
Theorem 100 (Product of compact spaces is compact)
If \(X\) and \(Y\) are compact topological spaces, then \(X \times Y\) is compact.
Proof: Product of compact spaces is compact
Suppose \(X\) and \(Y\) are compact topological spaces. Let \(\mathcal{V}\) be an open cover of \(X \times Y.\)
First, we consider the special case where each \(U \in \mathcal{V}\) has the form \(U = V \times W\) where \(V \in X\) and \(W \in Y.\) Since \(\mathcal{V}\) is an open cover of \(X \times Y,\) for each \((x, y) \in X \times Y\) there exists \(U_{xy} \in \mathcal{V}\) such that \((x, y) \in U_{xy}.\) By our previous assumption we can write this as \(U_{xy} = V_{xy} \times W_{xy}\) where \(V_{xy} \in X\) and \(W_{xy} \in Y.\)
Now fix \(x \in X.\) The set \(\{W_{xw}: y \in Y\}\) is an open cover of \(Y\) and since \(Y\) is compact, it has a finite subcover \(\{W_{xy_1}, \dots, W_{xy_N}\}.\) Now, define \(V_x = \cap_{n=1}^N V_{xy_n},\) and note that \(V_x\) is a finite intersection of open sets so it is open in \(X.\) In addition, \(\mathcal{U}_x = \{R_{xy_1}, \dots, U_{xy_N}\}\) is a finite open cover of \(V_x \times Y.\) Now, \(\{V_x: x \in X\}\) is an open cover of \(X\) and since \(X\) is compact, it has a finite subcover \(\{V_{x_1}, \dots, V_{x_M}\}.\) Therefore \(\cup_{m = 1}^M \mathcal{R}_{x_m}\) is a finite subset of \(\mathcal{V}\) which covers \(X \times Y.\)
Now consider the general case where \(\mathcal{V}\) is an arbitrary open cover of \(X \times Y.\) For each \((x, y) \in X \times Y,\) there exists open \(U_{xy} \in \mathcal{V}\) such that \((x, y) \in X \times Y.\) Since \(U_{xy}\) is open, there exist open \(V_{xy} \subseteq X\) and open \(W_{xy} \subseteq Y\) such that \(x \in V_{xy}\) and \(y \in W_{xy}\) and \(V_{xy} \times W_{xy} \subseteq U_{xy}.\) Then, the set \(\{V_{xy} \times W_{xy} : (x, y) \in X \times Y\}\) is an open cover of \(X \times Y\) of the special case we ahve considered above. Therefore it has a finite subcover \(\{V_{x_1y_1} \times W_{x_1y_1}, \dots, V_{x_Ny_N} \times W_{x_Ny_N}\}.\) Noting that \(V_{x_ny_n} \times W_{x_ny_n} \subseteq U_{x_ny_n},\) we see that \(\{U_{x_1y_1} \dots U_{x_Ny_N}\}\) has a finite subcover of \(X \times Y.\) Therefore \(X \times Y\) is compact.
Quotient spaces#
First, we remark that the quotient space of a compact space must be compact, since any open set in the quotient space can be related to an open set in the original space.
Lemma 41 (Quotient space of a compact space is compact)
Suppose \(X\) is a compact space and \(\sim\) an equivalence relation on \(X.\) The quotient space \(X / \sim\) is compact.
Proof: quotient space of a compact space is compact
Suppose \(X\) is a compact space and \(\sim\) an equivalence relation on \(X.\) Let \(\pi: X \to X / \sim\) be the projection map of \(\sim.\) Let \(\mathcal{V}\) be an open cover of \(X / \sim.\) Then for each \(V \in \mathcal{V},\) \(\pi^{-1}(V)\) is open in \(X,\) and the set \(\mathcal{W} = \{\pi^{-1}(V) \subseteq X: V \in \mathcal{V}\}\) is an open cover of \(X.\) Since \(X\) is compact \(\mathcal{W}\) has a finite subcover \(\{W_1, \dots, W_n\}\) so \(\{\pi(W_1), \dots, \pi(W_n)\} \subseteq \mathcal{V}\) is a finite open cover of \(X / \sim.\) Therefore \(X / \sim\) is compact.
We now show two useful results on quotient spaces, specifically we show two sufficient conditions for a map on a compact space to be a homeomorphism.
Lemma 42 (Sufficient condition for homeomorphism)
Suppose \(f: X \to Y\) is a continuous bijection. If \(X\) is compact and \(Y\) is Hausdorff, then \(f\) is a homeomorphism.
Proof: Sufficient condition for homeomorphism
By assumption, \(f\) is a continuous bijection so it remains to show that \(f^{-1}\) is continuous. We will show that if \(C \subseteq X\) is closed in \(X,\) then \((f^{-1})^{-1}(C)\) is closed in \(Y.\)
Since \(f\) is a bijection, \((f^{-1})^{-1}(C) = f(C).\) Since \(X\) is compact, \(C\) is compact which, together with the fact that \(f\) is continuous implies that \(f(C) = \text{im}f|_C\) is compact (Lemma 39). The set \(f(C) \subseteq Y\) is therefore a compact subspace of a Hausdorff space, so it is closed (Lemma 36).
Lemma 43 (Condition for homeomorphism for quotient spaces)
Suppose \(f: X / \sim~\to Y\) is a bijection, \(X\) is compact, \(Y\) is Hausdorff and \(f \circ \pi\) is continuous. Then \(f\) is a homeomorphism.
Proof: Condition for homeomorphism for quotient spaces
Note that the function \(\pi: X \to X / \sim\) which maps each element of \(X\) to its equivalence class in \(X / \sim,\) is continuous. Since \(X\) is compact and \(\pi\) is continuous, \(\text{im} \pi\) is compact (Lemma 39). Note that both \(\pi\) and \(\pi^{-1}\) are continuous. Since \(f \circ \pi\) is continuous, \(f = f \circ \pi \circ \pi^{-1}\) is continuous. Finally, since \(f\) is a continuous bijection, \(X\) is compact and \(Y\) is Hausdorff, \(f\) is a homeomorphism (Lemma 42).
Sequential compactness#
We now introduce another notion of compactness, namely sequential compactness.
Definition 110 (Sequential compactness)
A topological space \(X\) is sequentially compact if every sequence in \(X\) has a convergent subsequence.
Roughly speaking, sequential compacness captures the notion that a sequence cannot spread out arbitrarily in a space without having a subsequence that piles up around some point in the space. In general, compactness and sequential compactness are different properties, however in a metric space they are equivalent. To show this, we use the following intermediate lemma.
Lemma 44 (Equivalent condition for convergent subsequence in a metric space)
Let \((x_n)\) be a sequence in a metric space \((X, d)\) and \(x \in X.\) Then \((x_n)\) has a subsequence converging to \(x\) if and only if for every \(\epsilon > 0\) we have \(x_n \in B_\epsilon(x)\) for infinitely many \(n \in \mathbb{N}.\)
Proof: Equivalent condition for convergent sequence in a metric space
Let \((x_n)\) be a sequence in a metric space \((X, d)\) and \(x \in X.\)
Implies: Suppose \((x_n)\) has a subsequence \((x_{n_k})\) converging to \(x.\) Let \(\epsilon > 0.\) By the definition of convergence, there exists \(N \in \mathbb{N}\) such that for all \(m \geq N,\) we have \(x_m \in B_\epsilon(x).\) Therefore, for all \(n_k \geq m,\) of which there are infinitely many, we have \(x_{n_k} \in B_\epsilon(x).\)
Is implied by: Suppose that for every \(\epsilon > 0\) we have \(x_n \in B_\epsilon(x)\) for infinitely many \(n \in \mathbb{N}.\) We want to show \((x_n)\) has a subsequence converging to \(x.\) By hypothesis, \(x_n \in B_{1 / n}(x)\) for infinitely many \(n.\) For \(k = 1, 2, \dots,\) let \(n_k\) be the smallest such \(n\) with \(n_k > n_{k - 1},\) where we define \(n_0 = 0.\) Since \(d(x_{n_k}, x) < \frac{1}{n},\) we have that \(x_{n_k} \to x.\)
We are now ready to show that compactness and sequential compactness are equivalent in metric spaces.
Theorem 101 (Compact metric space \(\iff\) sequentially compact metric space)
A metric space \((X, d)\) is a compact if and only if it is sequentially compact.
Proof: Compact metric space \(~\iff\) sequentially compact metric space
Let \((X, d)\) be a metric space.
Implies: Suppose \((X, d)\) is compact. Let \((x_n)\) be a sequence in \(X\) and suppose that it does not have a convergent subsequence. Then, for any \(y \in X,\) there is no subsequence converging to \(y.\) By Lemma 44, there exists \(\epsilon_y > 0\) such that \(x_n \in B_\epsilon_y(y)\) only finitely many times. Let \(U_y = B_\epsilon_y(y).\) Now, the set \(\mathcal{V} = \{U_y: y \in X\}\) is an open cover of \(X.\) Since \(X\) is compact, there is a finite subcover \(\{U_{y_1}, \dots, U_{y_m}\}.\) Then \(x_n \cup_{i = 1}^m U_{y_i} = X\) only for finitely many \(n,\) but this is a contradiction because \(x_n \in X\) for all \(n.\) Therefore \((x_n)\) has a convergent subsequence and \((X, d).\)
Is implied by: Suppose \((X, d)\) is sequentially compact. Let \(\mathcal{V}\) be an open cover of \(X\) and suppose that there does not exist a finite open cover of \(X\) from it.
First, we show that there exists \(\delta > 0\) such that any open ball with radius less than \(\delta\) is contained within some \(V \in \mathcal{V}.\) This intermediate result is also known as Lebesgue’s number lemma. Suppose this is not the case. Then, for each \(n \in \mathcal{N},\) there exists \(x_n \in X\) such that \(B_{1 / n}(x_n)\) is not contained in any \(V \in \mathcal{V}.\) Since \(X\) is sequentially compact, the sequence \((x_n)\) has a convergent subsequence \((x_{n_k})\) with a limit \(x \in X.\) However, this leads to a contradiction. First, \(x\) cannot be in any \(V \in \mathcal{V},\) because if \(x\) is in some \(V \in \mathcal{V},\) then because \(V\) is open there exists \(r > 0\) such that \(B_r(x) \subseteq V. Second, this means that all \)x_{n_k}\( with \)n_k\( sufficiently large such that \)d(x_{n_k}, x) < r / 2\( and \)n_k > 2 / r\( for all \)n_k,\( would then satisfy \)B_{1 / n_k}(x_{n_k}) \subseteq B_r(x) \subseteq V.$ This is a contradiction, so the intermediate result holds.
Now let \(\delta > 0\) be such that any open ball with radius less than \(\delta\) is contained within some \(V \in \mathcal{V}.\) Pick \(x_1 \in X.\) The ball \(B_\delta(x_1)\) is contained in some \(V_1 \in \mathcal{V},\) and since \(V_1\) does not cover \(X\) there exists \(x_2 \in X \setminus B_\delta(x_1).\) We proceed recursively to obtain a sequence \((x_n)\) in this manner, which satisfies \(x_n \in X \setminus \cup_{i = 1}^{n - 1} B_\delta(x_i).\) The sequence \((x_{n_k})\) has a convergent subsequence, and since it converges it is Cauchy. However, this is a contradiction because no two elements \(x_{n_k}\) can belong to the same open ball of radius \(\delta.\) This contradicts the assumption that \(\mathcal{V}\) has no finite subcover so \(X\) is compact.
Completeness
We finish with a last section on completeness. This concerns metric spaces, but relies on results on compactness we have shown in this section. To define completeness, we first need to define Cauchy sequences.
Definition 111 (Cauchy sequence)
Let \((X, d)\) be a metric space. A sequence \((x_n)\) in \(X\) is Cauchy if for every \(\epsilon > 0,\) there exists \(N \in \mathbb{N}\) such that \(d(x_n, x_m) < \epsilon\) for all \(n, m \geq N.\)
A space is complete if every Cauchy sequence within it converges (to some limit in the space). Intutively, a Cauchy sequence is one whose the terms are progressively closer to one another. Completeness therefore means that all such sequences have a limit in the space.
Definition 112 (Complete metric space)
A metric space \((X, d)\) is complete if every Cauchy sequence in it converges.
For example \((0, 1)\) is a metric space that is not complete because the sequence \((a_n)\) where \(a_n = n^{-1}\) is Cauchy but does not converge to a point in \((0, 1).\) By contrast, we will show that \([0, 1]\) is complete, and this is a consequence of the space being compact, as shown by the following result.
Lemma 45 (Compact metric spaces are complete)
If the metric space \((X, d)\) is compact, it is also complete.
Proof: Compact metric spaces are complete
Suppose \((X, d)\) is a compact metric space. Let \((x_n)\) be a Cauchy sequence in \(X.\) From Theorem 101, it is sequentially compact. Since the space is sequentially compact, \((x_n)\) has a convergent subsequence \((x_{n_k}),\) which converges to some \(x \in X.\)
Let \(\epsilon > 0.\) Since \((x_n)\) is Cauchy, there exists \(N \in \mathbb{N}\) such that for all \(n, m \geq N,\) we have \(d(x_n, x_m) < \epsilon / 2.\) Since \((x_{n_k})\) converges to some \(x,\) there exists \(M \in \mathbb{N}\) such that for all \(n_k \geq M,\) we have \(d(x_{n_k}, x) < \epsilon / 2.\) Let \(l, n_k \geq \max\{N, M\}.\) Then
so \(x_l\) converges to \(x,\) and \(X\) is complete.
The interval \([0, 1]\) is a compact set so it is complete (Lemma 45. We conclude with a satisfying result that we can show with the ideas developed in the course: \(\mathbb{R}^n\) is complete.
Lemma 46 (\(\mathbb{R}^n\) is complete)
\(\mathbb{R}^n\) with the standard topology is complete.
Proof: \(~\mathbb{R}^n~\) is complete
If \((x_n)\) is a Cauchy sequence in \(\mathbb{R}^n,\) then \((x_n) \subseteq \overline{B}_r(0)\) for some \(r \in \mathbb{R}.\) Since \(\overline{B}_r(0)\) is a closed and bounded subset of a metric space, it is compact (Lemma 35). Since \((x_n)\) is a Cauchy sequence in a compact metric space, it converges (Lemma 45). Therefore \(\mathbb{R}^n\) is complete.