description/proof of that for sequence of metric spaces with induced topologies, this metric for product set induces product topology
Topics
About: topological space
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of product topology.
- The reader admits the proposition that for any metric space with the induced topology and any positive real number, the distance as the minimum of the original distance and the number is a metric and induces the original topology.
- The reader admits the proposition that any map from any topological space into any product topological space is continuous if and only if each component map is continuous.
Target Context
- The reader will have a description and a proof of the proposition that for any sequence of metric spaces with the induced topologies, this metric for the product set induces the product topology.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\{M_j \in \{\text{ the metric spaces }\}, \text{ with the metric }, dist'_j, and \text{ with the induced topology } \vert j \in \mathbb{N} \setminus \{0\}\}\):
\(\times_{j \in \mathbb{N} \setminus \{0\}} M_j\): \(= \text{ the product topological space }\) with the topology, \(O\)
\(r\): \(\in \mathbb{R}\), such that \(0 \lt r\)
\(\{dist_j = Min (\{dist'_j, r\}) \vert j \in \mathbb{N} \setminus \{0\}\}\):
\(r'\): \(\in \mathbb{R}\), such that \(1 / 2 + \sqrt{5} / 2 \lt r'\)
\(dist\): \(: (\times_{j \in \mathbb{N} \setminus \{0\}} M_j) \times (\times_{j \in \mathbb{N} \setminus \{0\}} M_j) \to \mathbb{R}, (m, m') \mapsto \sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m^j, m'^j) / r'^j\)
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Statements:
\(dist \in \{\text{ the metrics for } \times_{j \in \mathbb{N} \setminus \{0\}} M_j\}\)
\(\land\)
\(\text{ the topology induced by } dist, O' = O\)
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2: Note
\(1 / 2 + \sqrt{5} / 2 \lt r'\) is for guaranteeing that \(0 \lt r'^2 - r' - 1\), which Proof uses: as \(1 / 2 + \sqrt{5} / 2 \lt r'\), \(\sqrt{5} / 2 \lt r' - 1 / 2\), \(5 / 4 \lt (r' - 1 / 2)^2\), and \(0 \lt (r' - 1 / 2)^2 - 5 / 4 = r'^2 - r' - 1\).
\(1 \lt r'\), because \(1 \lt \sqrt{5} / 2\).
\(dist\) is not the only metric but is a metric that induces \(\times_{j \in \mathbb{N} \setminus \{0\}} M_j\).
While the proposition that for any finite-product metric space, the topology induced by the product metric is the product topology of the topologies induced by the constituent metrics has been proved, this proposition is a kind of substitution for when it is the product of a sequence of metric spaces.
3: Proof
Whole Strategy: Step 1: see that \(dist\) is a metric; Step 2: take \(id: (\times_{j \in \mathbb{N} \setminus \{0\}} M_j)_{O'} \to \times_{j \in \mathbb{N} \setminus \{0\}} M_j\) where the domain is with \(O'\); Step 3: see that \(id\) is continuous; Step 4: see that \(id^{-1}\) is continuous; Step 5: conclude the proposition.
Step 1:
Let us see that \(dist\) is a metric.
\(\sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m^j, m'^j) / r'^j \in \mathbb{R}\), because \(\sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m^j, m'^j) / r'^j \le \sum_{j \in \mathbb{N} \setminus \{0\}} r / r'^j = r \sum_{j \in \mathbb{N} \setminus \{0\}} 1 / r'^j = r (1 / r') / (1 - 1 / r')\).
\(dist_j\) is a metric on \(M_j\), which induces the topology induced by \(dist'_j\), by the proposition that for any metric space with the induced topology and any positive real number, the distance as the minimum of the original distance and the number is a metric and induces the original topology.
For each \(m, m', m'' \in \times_{j \in \mathbb{N} \setminus \{0\}} M_j\), 1) \(0 \le dist (m, m')\) and \(dist (m, m') = 0 \iff m = m'\): when \(dist (m, m') = 0\), \(dist_j (m^j, m'^j) = 0\) for each \(j\), so, \(m^j = m'^j\) for each \(j\), so, \(m = m'\); when \(m = m'\), \(m^j = m'^j\) for each \(j\), so, \(dist_j (m^j, m'^j) = 0\) for each \(j\), so, \(dist (m, m') = 0\); 2) \(dist (m, m') = dist (m', m)\): \(dist (m, m') = \sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m^j, m'^j) / r'^j = \sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m'^j, m^j) / r'^j = dist (m', m)\); 3) \(dist (m, m'') \le dist (m, m') + dist (m', m'')\): \(dist (m, m'') = \sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m^j, m''^j) / r'^j \le \sum_{j \in \mathbb{N} \setminus \{0\}} (dist_j (m^j, m'^j) + dist_j (m'^j, m''^j)) = \sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m^j, m'^j) / r'^j + \sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m'^j, m''^j) / r'^j = dist (m, m') + dist (m', m'')\).
So, \(dist\) is a metric for \(M\).
Step 2:
Let \((\times_{j \in \mathbb{N} \setminus \{0\}} M_j)_{O'}\) be the product set, \(\times_{j \in \mathbb{N} \setminus \{0\}} M_j\), with the topology, \(O'\).
Let \(id: (\times_{j \in \mathbb{N} \setminus \{0\}} M_j)_{O'} \to \times_{j \in \mathbb{N} \setminus \{0\}} M_j\) be the identity.
We are going to see that \(id\) is a homeomorphism, because if so, for each \(U' \in O'\), \(id (U') = U' \in O\), and for each \(U \in O\), \(id^{-1} (U) = U \in O'\), so, \(O = O'\) will hold.
Step 3:
Let us see that \(id\) is continuous.
For each \(j \in \mathbb{N} \setminus \{0\}\), let \(\pi^j: \times_{j \in \mathbb{N} \setminus \{0\}} M_j \to M_j\) be the projection.
Let us see that \(\pi^j \circ id: (\times_{j \in \mathbb{N} \setminus \{0\}} M_j)_{O'} \to M_j\) is continuous.
Let \(m \in (\times_{j \in \mathbb{N} \setminus \{0\}} M_j)_{O'}\) be any.
\(\pi^j \circ id (m) = m^j\), and let \(U_{m^j} \subseteq M_j\) be any open neighborhood of \(m^j\).
There is a \(B_{m^j, \epsilon} \subseteq M_j\), where \(B_{m^j, \epsilon}\) is by \(dist_j\), such that \(B_{m^j, \epsilon} \subseteq U_{m^j}\), because \(M_j\) has the topology induced by \(dist_j\).
Let \(B'_{m, \epsilon / r'^j} \subseteq (\times_{j \in \mathbb{N} \setminus \{0\}} M_j)_{O'}\) be by \(dist\).
For each \(b' \in B'_{m, \epsilon / r'^j}\), \(dist (m, b') = \sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m^j, b'^j) / r'^j \lt \epsilon / r'^j\).
So, \(dist_j (m^j, b'^j) / r'^j \le \sum_{j \in \mathbb{N} \setminus \{0\}} dist_j (m^j, b'^j) / r'^j \lt \epsilon / r'^j\), so, \(dist_j (m^j, b'^j) \lt \epsilon\).
That means that \(\pi^j \circ id (b') \in B_{m^j, \epsilon}\), so, \(\pi^j \circ id (B'_{m, \epsilon / r'^j}) \subseteq B_{m^j, \epsilon}\).
\(B'_{m, \epsilon / r'^j}\) is an open neighborhood of \(m\) on \((\times_{j \in \mathbb{N} \setminus \{0\}} M_j)_{O'}\), because \(O'\) is induced by \(dist\).
So, \(\pi^j \circ id\) is continuous.
\(id\) is continuous, by the proposition that any map from any topological space into any product topological space is continuous if and only if each component map is continuous.
Step 4:
Let us see that \(id^{-1}\) is continuous.
Let \(m \in \times_{j \in \mathbb{N} \setminus \{0\}} M_j\) be any.
Let \(U'_m \subseteq (\times_{j \in \mathbb{N} \setminus \{0\}} M_j)_{O'}\) be any open neighborhood of \(id^{-1} (m) = m\).
There is a \(B'_{m, \epsilon} \subseteq (\times_{j \in \mathbb{N} \setminus \{0\}} M_j)_{O'}\), where \(B'_{m, \epsilon}\) is by \(dist\), such that \(B'_{m, \epsilon} \subseteq U'_m\), because \(O'\) is induced by \(dist\).
Let us take \(n \in \mathbb{N} \setminus \{0\}\) such that \(1 / r'^n \lt \epsilon (1 - 1 / r' - 1 / r'^2) / r\): \(0 \lt r'^2 - r' - 1\) as is mentioned in Note, so, \(0 \lt 1 - 1 / r' - 1 / r'^2\), so, such an \(n\) exists.
Let us take \(B_{m^1, \epsilon / r'} \times ... \times B_{m^{n - 1}, \epsilon / r'} \times M_n \times M_{n + 1} \times ... \subseteq \times_{j \in \mathbb{N} \setminus \{0\}} M_j\), where \(B_{m^j, \epsilon / r'}\) is by \(dist_j\), which is an open neighborhood of \(m\), because \(B_{m^j, \epsilon / r'} \subseteq M_j\) is open and only finite of the constituents are not \(M_j\) s.
Let \(m' \in B_{m^1, \epsilon / r'} \times ... \times B_{m^{n - 1}, \epsilon / r'} \times M_n \times M_{n + 1} \times ...\) be any.
\(dist (m, m') = \sum_{j \in \{1, ..., n - 1\}} dist_j (m^j, m'^j) / r'^j + \sum_{j \in \mathbb{N} \setminus \{0, 1, ..., n - 1\}} dist_j (m^j, m'^j) / r'^j \lt \sum_{j \in \{1, ..., n - 1\}} \epsilon / r' / r'^j + \sum_{j \in \mathbb{N} \setminus \{0, 1, ..., n - 1\}} r / r'^j \lt \epsilon / r' \sum_{j \in \mathbb{N} \setminus \{0\}} 1 / r'^j + r / r'^{n - 1} \sum_{j \in \mathbb{N} \setminus \{0\}} 1 / r'^j = (\epsilon / r' + r / r'^{n - 1}) \sum_{j \in \mathbb{N} \setminus \{0\}} 1 / r'^j = (\epsilon / r' + r / r'^{n - 1}) 1 / r' / (1 - 1 / r') = (\epsilon / r'^2 + r / r'^n) / (1 - 1 / r') \lt (\epsilon / r'^2 + r \epsilon (1 - 1 / r' - 1 / r'^2) / r) / (1 - 1 / r') = \epsilon (1 / r'^2 + 1 - 1 / r' - 1 / r'^2) / (1 - 1 / r') = \epsilon (1 - 1 / r') / (1 - 1 / r') = \epsilon\).
So, \(id^{-1} (m') = m' \in B'_{m, \epsilon}\).
So, \(id^{-1} (B_{m^1, \epsilon / r'} \times ... \times B_{m^{n - 1}, \epsilon / r'} \times M_n \times M_{n + 1} \times ...) \subseteq B'_{m, \epsilon}\).
So, \(id^{-1}\) is continuous.
Step 5:
So, \(id\) is a homeomorphism.
So, \(O' = O\).
