description/proof of that union of topologist's sine curve and interval of y-axis around \(0\) is connected but not path-connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological subspace.
- The reader knows a definition of connected topological space.
- The reader knows a definition of path-connected topological space.
- The reader admits the proposition that any map from any topological space into any finite product topological space is continuous if and only if all the component maps are continuous.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
- The reader admits the proposition that the set of all the connected topological subspaces of the \(\mathbb{R}\) Euclidean topological space is the set of all the intervals.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of any connected subspace is connected.
- The reader admits the proposition that the product of any connected topological spaces is connected.
- 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 possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
- The reader admits the proposition that for any topological space, the union of any 2 connected subspaces is connected if each neighborhood of a point on one of the subspaces contains a point of the other subspace.
Target Context
- The reader will have a description and a proof of the proposition that the union of the topologist's sine curve and any interval of the y-axis around \(0\) is connected but not path-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:
\(J\): \(\in \{\text{ the intervals of } \mathbb{R} \text{ around } 0\}\), \(= (r_1, r_2), (r_1, r_2], [r_1, r_2), \text{ or } [r_1, r_2]\)
\(T\): \(\in \{\text{ the topological spaces }\}\), \(= \{(x, sin (1 / x)) \vert 0 \lt x\} \cup (\{0\} \times J) \subseteq \mathbb{R}^2\), with the subspace topology
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Statements:
\(T \in \{\text{ the connected topological spaces }\}\)
\(\land\)
\(T \notin \{\text{ the path-connected topological spaces }\}\)
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2: Note
\(\{(x, sin (1 / x)) \vert 0 \lt x\} \subseteq \mathbb{R}^2\) is called "topologist's sine curve".
Or \(\{(x, sin (1 / x)) \vert 0 \lt x \lt 1\} \subseteq \mathbb{R}^2\) may be sometimes used, but the difference does not influence our purpose.
3: Proof
Whole Strategy: Step 1: see that \(\{(x, sin (1 / x)) \vert 0 \lt x\} \subseteq \mathbb{R}^2\) is connected; Step 2: see that \(\{0\} \times J \subseteq \mathbb{R}^2\) is connected; Step 3: see that \(T\) is connected; Step 4: see that \(T\) is not path-connected, by seeing that \((0, 0)\) and \((1 / \pi, 0)\) are not path-connected.
Step 1:
\(f: (0, \infty) \to \mathbb{R}^2, x \mapsto (x, sin (1 / x))\) is continuous, by the proposition that any map from any topological space into any finite product topological space is continuous if and only if all the component maps are continuous: \(: x \mapsto x\) is continuous; \(x \mapsto sin (1 / x)\) is continuous, because \(: x \mapsto 1 / x\) and the sine map is continuous and the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point applies.
\((0, \infty)\) is connected, by the proposition that the set of all the connected topological subspaces of the \(\mathbb{R}\) Euclidean topological space is the set of all the intervals.
\(f ((0, \infty)) = \{(x, sin (1 / x)) \vert 0 \lt x\} \subseteq \mathbb{R}^2\) is connected, by the proposition that for any continuous map between any topological spaces, the image of any connected subspace is connected.
Step 2:
\(\{0\} \subseteq \mathbb{R}\) is obviously connected.
\(J \subseteq \mathbb{R}\) is connected, by the proposition that the set of all the connected topological subspaces of the \(\mathbb{R}\) Euclidean topological space is the set of all the intervals.
\(\{0\} \times J \subseteq \mathbb{R}^2\) is connected, by the proposition that the product of any connected topological spaces is connected, 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 \(d\), and the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
Step 3:
Let \(U_{0, 0} \subseteq \mathbb{R}^2\) be any open neighborhood of \((0, 0)\) on \(\mathbb{R}^2\).
There is a \(B_{(0, 0), \epsilon} \subseteq \mathbb{R}^2\) such that \(B_{(0, 0), \epsilon} \subseteq U_{0, 0}\).
There is an \(0 \lt x \lt \epsilon\) such that \(1 / x = n \pi\) where \(n \in \mathbb{N} \setminus \{0\}\).
\((x, sin (1 / x)) = (x, 0) \in B_{(0, 0), \epsilon} \subseteq U_{0, 0}\).
By the proposition that for any topological space, the union of any 2 connected subspaces is connected if each neighborhood of a point on one of the subspaces contains a point of the other subspace, \(T\) is connected.
Step 4:
Let us see that \(T\) is not path-connected, by seeing that \((1 / \pi, 0)\) and \((0, 0)\) are not path-connected.
Let us suppose that there was a path, \(\gamma: [0, 1] \to T\), such that \(\gamma (0) = (1 / \pi, 0)\) and \(\gamma (1) = (0, 0)\).
Let \(\pi^1: \mathbb{R}^2 \to \mathbb{R}\) be the projection to the 1st component, which would be obviously continuous.
\(\pi^1 \circ \gamma: [0, 1] \to \mathbb{R}\) would be continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
Let us take \(r := Inf (\{r' \in [0, 1] \vert \pi^1 \circ \gamma (r') = 0\})\), which would exist, because at least \(\pi^1 \circ \gamma (1) = 0\), so, \(\{r' \in [0, 1] \vert \pi^1 \circ \gamma (r') = 0\}\) would not be empty and would be lower bounded, and any nonempty lower bounded subset of \(\mathbb{R}\) would have the infimum, as was well known.
\(\pi^1 \circ \gamma (r) = 0\), because otherwise, there would be a \(B_{\pi^1 \circ \gamma (r), \epsilon} \subseteq \mathbb{R}\) where \(\epsilon \lt \pi^1 \circ \gamma (r)\) and a \(B_{r, \delta} \subseteq [0, 1]\) such that \(\pi^1 \circ \gamma (B_{r, \delta}) \subseteq B_{\pi^1 \circ \gamma (r), \epsilon}\), because \(\pi^1 \circ \gamma\) was continuous, and \(r\) would not be the infimum.
Let \(\gamma (r) = (0, a)\).
Let us take \(B_{(0, a), \epsilon} \subseteq \mathbb{R}^2\) where \(\epsilon \lt 1\).
There would be a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\), \([r - \delta, r] \subseteq [0, 1]\), and \(\gamma ([r - \delta, r]) \subseteq B_{(0, a), \epsilon}\), because \(\gamma\) was continuous.
\(0 \lt \pi^1 \circ \gamma (r - \delta)\), because \(r\) was the infimum.
\(\pi^1 \circ \gamma ([r - \delta, r])\) would be an interval that contained \([0, \pi^1 \circ \gamma (r - \delta)]\), by the proposition that for any continuous map between any topological spaces, the image of any connected subspace is connected and the proposition that the set of all the connected topological subspaces of the \(\mathbb{R}\) Euclidean topological space is the set of all the intervals.
So, there would be an \(t \in [r - \delta, r]\) such that \(\pi^1 \circ \gamma (t) \in (0, \pi^1 \circ \gamma (r - \delta))\) such that \(\pi^1 \circ \gamma (t) = 1 / (n \pi + 1 / 2 \pi) \text{ or } 1 / (n \pi + 3 / 2 \pi)\) where \(n \in \mathbb{N}\).
Then, \(\gamma (t) = (1 / (n \pi + 1 / 2 \pi), 1) \text{ or } (1 / (n \pi + 3 / 2 \pi), -1)\).
But whatever \(a\) was, not the both of \((1 / (n \pi + 1 / 2 \pi), 1)\) and \((1 / (n \pi + 3 / 2 \pi), -1)\) could be in \(B_{(0, a), \epsilon}\), because \(\epsilon \lt 1\): if \(a - \epsilon \lt 1 \lt a + \epsilon\), \(- 1 \lt 1 - 2 \epsilon \lt a - \epsilon\); if \(a - \epsilon \lt -1 \lt a + \epsilon\), \(a + \epsilon \lt -1 + 2 \epsilon \lt 1\).
That is a contradiction against that \(\gamma ([r - \delta, r]) \subseteq B_{(0, a), \epsilon}\).
So, there is no such \(\gamma\).
So, \((1 / (n \pi), 0)\) and \((0, 0)\) are not path-connected.
