James Munkres Topology: Lemma 21.2 The sequence lemma
Lemma 21.2 (The sequence lemma) Let \(X\) be a topological space; let \(A \subset X\). If there is a sequence of points of \(A\) converging to \(x\), then \(x \in \bar{A}\); the converse holds if \(X\) is metrizable.
Proof a) Sequence convergence \(\Longrightarrow\) the limit point belongs to \(\bar{A}\).
Let \(\{x_n\}_{n \in \mathbb{Z_+}}\) be a sequence of points in \(A\). When \(n \rightarrow \infty\), it converges to \(x\) topologically in \(X\). Then for all open set \(U\) containing \(x\), there exists an \(N \in \mathbb{Z_+}\), such that when \(n > N\), \(x_n \in U\). Hence \(U \cap A \neq \Phi\). According to Theorem 17.5 (a), \(x \in \bar{A}\).
b) \(x\) belongs to \(\bar{A}\) and \(X\) is metrizable \(\Longrightarrow\) Sequence convergence to \(x\).
Still according to Theorem 17.5 (a), when \(x \in \bar{A}\), for all open set \(U\) containing \(x\), \(U \cap A \neq \Phi\). However, this only ensures that the intersection is nonempty but is not enough to promise that there exists an \(N\) in \(\mathbb{Z}_+\), such that for all \(n > N\), \(x_n\) belongs to \(U\). Hence, the desired convergence sequence in \(A\) does not necessarily exist.
If \(X\) is assigned a metric, a collection of nested open balls \(\left\{B_n(x, \frac{1}{n})\right\}_{n \in \mathbb{Z}_+}\) centered at \(x \in \bar{A}\) can be constructed. For all \(n \in \mathbb{Z}_+\), \(B_{n}(x, \frac{1}{n}) \cap A \neq \Phi\) and an element \(x_n\) can be selected from this intersection. Thus a sequence \(\{x_n\}_{n \in \mathbb{Z}_+}\) convergent to \(x\) is obtained.
Remark
As shown in b) above, a metric assigned to the space \(X\) which generates the same topology as that used for defining sequence convergence is mandatory to ensure the existence of a convergent sequence to \(x\). This contradicts our common conception about the equivalence between the closeness of a set \(A\) and the existence of a convergent sequence with its limit point within \(A\). This is because the spaces we are dealing with in everyday life, such as Banach spaces, Hilbert spaces, have sound properties which have already included a well defined metric. However, when we come to the study of topology, such nice property is stripped away for the purpose of establishing a more abstract and general theory underpinning those high level and realistic topics.
It is natural for us to ask that if an example can be given, where the space \(X\) has no associated metric and there is no sequence \(\{x_n\}_{n \in \mathbb{Z}_+}\) in \(A\) convergent to a point \(x \in \bar{A}\).
Let’s consider a set \(X = S_{\Omega} \cup \{\Omega\}\) with \(S_{\Omega}\) being the minimal uncountable well-ordered set as defined in Lemma 10.2. Let \(X\) be assigned the order topology and let \(A = S_{\Omega}\) be a subset of \(X\). Because \(S_{\Omega}\) is the largest element in \(X\), any open set \(U\) in \(X\) containing \(\Omega\) must have the form \((x ,\Omega]\) with \(x \in S_{\Omega}\). Then it is obvious that \(U \cap S_{\Omega} \neq \Phi\) and thus \(\Omega\) belongs to \(\bar{S}_{\Omega}\). More accurately speaking, \(\Omega\) is a limit point of \(S_{\Omega}\).
Next, we show that there is no sequence \(\{x_n\}_{n \in \mathbb{Z}_+}\) in \(S_{\Omega}\) convergent to \(\Omega\).
Assume such sequence really exists. Because it is a countable set, according to Theorem 10.3, it has an upper bound \(x^*\) in \(S_{\Omega}\). We know from Lemma 10.2 that \(S_{\Omega}\) is uncountable and the section \(S_{x^*}\) is countable, therefore the set \(V = \{x | x \in S_{\Omega} \;\text{and}\; x > x^* \}\) is not empty. Then the open set \((x^*, \Omega]\) in \(X\) containing \(\Omega\) has an empty intersection with the sequence \(\{x_n\}_{n \in \mathbb{Z}_+}\) in \(S_{\Omega}\). Therefore, \(\{x_n\}_{n \in \mathbb{Z}_+}\) is not convergent to \(\Omega\).
The contrapositive of part b) in this lemma can be used to prove that a space with a certain topology is not metrizable, i.e. by showing that there exists an abnormal point \(x\) in the closure of \(A\), to which there is no convergent sequence in \(A\), we can prove that there is no metric for the space \(X\) which can induce the same topology as that used for defining the sequence convergence.
For all \(A \subset X\) and for all \(x \in \bar{A}\), if there is always a sequence in \(A\) convergent to \(x\), we still cannot assert that \(X\) is metrizable.
This can be verified by giving a counter example. Let \(X = \mathbb{R}\) be given the finite complement topology, i.e. the space satisfies the \(T_1\) axiom. Then for any \(A \subset X\), if \(A\) is a finite set, \(A\) itself is closed. For all \(x \in \bar{A} = A\), \(\{x_n = x\}_{n \in \mathbb{Z}_+}\) is a sequence in \(A\) convergent to \(x\).
If \(A\) is an infinite set, \(\bar{A} = \mathbb{R}\). This is because for all \(x \in \mathbb{R}\) and for all open set \(U\) in \(\mathbb{R}\) containing \(x\), its complement \(U^{\rm c}\) is closed and is thus finite. Assume \(U \cap A = \Phi\), then \(A \subset U^{\rm c}\). However, because \(A\) is infinite, it cannot be contained within the finite set \(U^{\rm c}\). Therefore, \(U \cap A \neq \Phi\) and it proves that for all \(x \in \mathbb{R}\) it belongs to the closure of \(A\). Hence \(\bar{A} = \mathbb{R}\).
Let \(\{x_n\}_{n \in \mathbb{Z}_+}\) be a sequence in \(A\) which has an infinite number of different elements. This is feasible because \(A\) itself is an infinite set. For all \(x \in \mathbb{R}\) and for all open set \(U\) in \(\mathbb{R}\) containing \(x\), its complement \(U^{\rm c}\) is a closed finite set. Then we consider following two complete cases.
- If for all \(y \in U^{\rm c}\), \(y \notin \{x_n\}_{n \in \mathbb{Z}_+}\), we have \(\{x_n\}_{n \in \mathbb{Z}_+} \subset U\), i.e. in the language of convergence, for all \(n >1 \), \(x_n \in U\).
- If there exists a finite subset \(V\) of \(U^{\rm c}\) such that \(V \subset \{x_n\}_{n \in \mathbb{Z}_+}\) and let \(N\) be the maximum index in the sequence for those elements in \(V\), then for all \(n > N\), \(x_n \in U\).
Therefore, the sequence \(\{x_n\}_{n \in \mathbb{Z}_+}\) converges to \(x\) in any of the above two cases.
This conclusion can be restated as below.
Let \(\mathbb{R}\) be assigned the finite complement topology. Any sequence \(\{x_n\}_{n \in \mathbb{Z}_+} \subset \mathbb{R}\) having an infinite number of different elements can converge to any point \(x\) in \(\mathbb{R}\).
Next, we need a small lemma to be proved:
Every topological space \(X\) with a metric \(d\) satisfies the Hausdorff axiom.
Proof For all \(x, y \in X\), let their distance be \(d(x, y) = \epsilon\). Select an open ball \(B_d(x, \frac{\epsilon}{2})\) and for all \(z \in B_d(x, \frac{\epsilon}{2})\), we have \(d(x, y) \leq d(x, z) + d(z, y)\) and thus \(d(z, y) \geq d(x, y) - d(x, z)\). Because \(d(x, z) < \frac{\epsilon}{2}\), \(d(z, y) > \epsilon - \frac{\epsilon}{2} = \frac{\epsilon}{2}\). Hence \(z \notin B_d(y, \frac{\epsilon}{2})\). Similarly, for all \(z \in B_d(y, \frac{\epsilon}{2})\), \(z \notin B_d(x, \frac{\epsilon}{2})\). Therefore, \(X\) satisfies the Hausdorff axiom.
Up to now, the conditions in the proposition of this part of the remark have been met. Because \(\mathbb{R}\) with the finite complement topology only satisfies the \(T_1\) axiom, which is a weaker condition than the Hausdorff axiom, according to the contrapositive of the above lemma, \(\mathbb{R}\) is not metrizable.
