description/proof of that subspace of Euclidean topological space whose projection into component is non-single countable is not connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of topological subspace.
- The reader knows a definition of connected topological space.
- The reader admits the proposition that the \(d\)-dimensional Euclidean topological space is homeomorphic to the product of any combination of some lower-dimensional Euclidean spaces whose (the product's) dimension equals \(d\).
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
Target Context
- The reader will have a description and a proof of the proposition that any subspace of any Euclidean topological space whose projection into any component is non-single countable is not connected.
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:
\(\mathbb{R}^d\): \(= \text{ the Euclidean topological space }\)
\(T\): \(\in \{\text{ the topological subspaces of } \mathbb{R}^d \}\)
\(\pi_j\): \(: \mathbb{R}^d \to \mathbb{R}, (x^1, ..., x^d) \mapsto (x^j)\)
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Statements:
\(\pi_j (T) \in \{\text{ the countable sets }\} \land 1 \lt \vert \pi_j (T) \vert\)
\(\implies\)
\(T \notin \{\text{ the connected topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: take an \(\{r_1, r_2\} \subseteq \pi_j (T)\) such that \(r_1 \lt r_2\) and an \(r \in \mathbb{R}\) such that \(r_1 \lt r \lt r_2\) and \(r \notin \pi_j (T)\); Step 2: take the open subsets, \(U'_1 := \mathbb{R}^{j - 1} \times (- \infty, r) \times \mathbb{R}^{d - j}, U'_2 := \mathbb{R}^{j - 1} \times (r, \infty) \times \mathbb{R}^{d - j} \subseteq \mathbb{R}^d\), and see that \(T = (T \cap U'_1) \cup (T \cap U'_2)\); Step 3: see that \(T \cap U'_l\) is a nonempty open subset of \(T\) such that \((T \cap U'_1) \cap (T \cap U'_2) = \emptyset\).
Step 1:
Let us take an \(\{r_1, r_2\} \subseteq \pi_j (T)\) such that \(r_1 \lt r_2\), which is possible because \(1 \lt \vert \pi_j (T) \vert\).
There is an \(r \in \mathbb{R}\) such that \(r_1 \lt r \lt r_2\) and \(r \notin \pi_j (T)\), because \(\pi_j (T)\) is countable: the interval, \((r_1, r_2)\), has some uncountable points (as is well-known), and the countable \(\pi_j (T)\) cannot cover the whole \((r_1, r_2)\).
Step 2:
Let us take the subsets, \(U'_1 := \mathbb{R}^{j - 1} \times (- \infty, r) \times \mathbb{R}^{d - j}, U'_2 := \mathbb{R}^{j - 1} \times (r, \infty) \times \mathbb{R}^{d - j} \subseteq \mathbb{R}^d\).
They are open on \(\mathbb{R}^d\), by the proposition that the \(d\)-dimensional Euclidean topological space is homeomorphic to the product of any combination of some lower-dimensional Euclidean spaces whose (the product's) dimension equals \(d\): \((- \infty, r)\) and \((r, \infty)\) are open on \(\mathbb{R}\).
\(T = T \cap (U'_1 \cup U'_2)\), because for each \(t = (x^1, ..., x^j, ..., x^d) \in T\), \(x^j \in (- \infty, r)\) or \(x^j \in (r, \infty)\), so, \(t \in U'_1\) or \(t \in U'_2\).
\(= (T \cap U'_1) \cup (T \cap U'_2)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset
Step 3:
\(T \cap U'_1 \neq \emptyset\), because \(r_1 \in (- \infty, r)\); \(T \cap U'_2 \neq \emptyset\), likewise.
\(T \cap U'_1\) is open on \(T\), because \(U'_1\) is open on \(\mathbb{R}^d\), by the definition of subspace topology; \(T \cap U'_2\) is open on \(T\), likewise.
\((T \cap U'_1) \cap (T \cap U'_2) = \emptyset\), because for each \(t = (x^1, ..., x^j, ..., x^d) \in T \cap U'_1\), \(x^j \in (-\infty, r)\), so, \(x^j \notin (r, \infty)\), so, \(t \notin T \cap U'_2\).
So, \(T\) is not connected.
