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A description/proof of that pair of open sets of connected topological space is finite-open-sets-sequence-connected
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any pair of open sets of any connected topological space is finite-open-sets-sequence-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: Description
For any connected topological space, \(T\), and any pair of open sets, \(U_1, U_2 \subseteq T\), \(U_1\) and \(U_2\) are finite-open-sets-sequence-connected.
2: Proof
Take any open cover, \(S_c = \{U_\alpha\}\), of \(T\) that includes \(U_1\) and \(U_2\), which is always possible because we can take the set of neighborhoods of all the point, with \(U_1\) and \(U_2\) added. Take the equivalence class, \(S_e = \{U_\beta\} \subseteq S_c\), of the open cover such that each element of the equivalence class is finite-open-sets-sequence-connected with \(U_1\) via some elements of \(S_c\), which is obviously an equivalence class.
If \(S_e\) did not equal \(S_c\), \(S_r := S_c \setminus S_e = \{U_\gamma\}\) would be nonempty. As \(S_e \cup S_r = S_c\) is an open cover, \((\cup_\beta U_\beta) \cup (\cup_\gamma U_\gamma) = T\), but as \(T\) is connected, \((\cup_\beta U_\beta) \cap (\cup_\gamma U_\gamma) \neq \emptyset\) because otherwise, \(T\) would be the disjoint union of open sets, \(\cup_\beta U_\beta\) and \(\cup_\gamma U_\gamma\). So, there would be a point, \(p \in T\), such that \(p \in \cup_\beta U_\beta\) and \(p \in \cup_\gamma U_\gamma\), which means that \(p \in U_\beta\) for an \(\beta\) and \(p \in U_\gamma\) for a \(\gamma\), so, \(p \in U_\beta \cap U_\gamma\), which is a contradiction, because as \(U_\gamma\) would share a point with \(U_\beta\), \(U_\gamma\) should be finite-open-sets-sequence-connected with \(U_1\). So, \(S_e\) equals \(S_c\).
So, \(U_2 \in S_e\), and \(U_1\) and \(U_2\) are finite-open-sets-sequence-connected.
3: Note
Although not every connected topological space is path-connected, any pair of open sets of any connected topological space is connected by way of open sets.
References
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A definition of finite-open-sets-sequence-connected pair of open sets
Topics
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topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of finite-open-sets-sequence-connected pair of open sets.
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: Definition
For any topological space, \(T\), any pair of open sets, \(U_1, U_2 \subseteq T\), such that there is a finite sequence of open sets of \(T\), \(U_1, U_3, . . ., U_n, U_2\), such that any adjoining pair of elements in the sequence shares a point.
References
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A description/proof of that closed discrete subspace of compact topological space has only finite points
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any closed discrete subspace of any compact topological space has only finite points.
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: Description
For any compact topological space, \(T\), and any closed discrete subspace, \(S \subseteq T\), \(S\) has only finite points.
2: Proof
Suppose that \(S\) had some infinite points. By the proposition that any compact topological space has an accumulation point of any subset with infinite points, \(T\) would have an accumulation point, \(p \in T\), of \(S\). But by the proposition that any topological subset is closed if and only it equals its closure, \(S = \overline{S}\), but by the proposition that the closure of any topological subset equals the union of the subset and the set of the accumulation points of the subset, the accumulation point would be contained in \(S\), which is impossible, because as \(S\) is discrete, each point is open, so also the accumulation point would be an open set by itself, which is against the definition of accumulation point, a contradiction.
References
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A description/proof of that compact topological space has accumulation point of subset with infinite points
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any compact topological space has an accumulation point of any subset with infinite points.
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: Description
For any compact topological space, \(T\), and any subset, \(S \subseteq T\), with infinite points, \(T\) has an accumulation point, \(p' \in T\) of \(S\).
2: Proof
Suppose that \(T\) had no accumulation point of \(S\). Then, around each point, \(p \in T\), there would be an open set, \(p \in U_p \subseteq T\), such that \(U_p \cap S = \emptyset \text{ or } {p}\). Such open sets would constitute an open cover of \(T\), which would have a finite subcover. As each open set would have only 1 point of \(S\) at most, \(S\) would have only finite points, which is a contradiction.
References
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A description/proof of that closure of difference of subsets is not necessarily difference of closures of subsets, but is contained in closure of minuend
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the closure of the difference of any 2 subsets is not necessarily the difference of the closures of the subsets, but is contained in the closure of the minuend.
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: Description
For any topological space, \(T\), and any subsets, \(S_1, S_2 \subseteq T\), the closure of the difference of the subsets, \(\overline{S_1 \setminus S_2}\), is not necessary the difference of the closures of the subsets, \(\overline{S_1} \setminus \overline{S_2}\), but is contained in the closure of the minuend, \(\overline{S_1}\), which means \(\overline{S_1 \setminus S_2} \subseteq \overline{S_1}\).
2: Proof
Suppose that \(T\) is the \(\mathbb{R}^2\) Euclidean topological space, \(S_1\) is the open ball, \(B_{0-2}\), and \(S_2\) is the open ball, \(B_{0-1}\), where \(B_{p-r}\) denotes the open ball centered at \(p\) with the \(r\) diameter. Then, \(\overline{S_1 \setminus S_2}\) contains the border of \(B_{0-1}\), but \(\overline{S_1} \setminus \overline{S_2}\) does not. So, as there is a counterexample, \(\overline{S_1 \setminus S_2}\) is not necessarily \(\overline{S_1} \setminus \overline{S_2}\).
\(S_1 \setminus S_2 = S_1 \cap (T \setminus S_2)\). By the proposition that the closure of the intersection of any finite number of subsets is contained in the intersection of the closures of the subsets, \(\overline{S_1 \setminus S_2} \subseteq \overline{S_1} \cap \overline{(T \setminus S_2)} \subseteq \overline{S_1}\).
References
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A description/proof of that for topological space, intersection of basis and subspace is basis for subspace
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any topological space, the intersection of any basis and any subspace is a basis for the subspace.
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: Description
For any topological space, \(T\), any basis, \(B = \{B_\alpha\}\), and any subset, \(S \subseteq T\), with the subspace topology, the intersection of the basis and the subset, \(\{B_\alpha \cap S\}\), is a basis for the subspace.
2: Proof
For any open set, \(U \subseteq S\), on \(S\), \(U = U' \cap S\) where \(U' \subseteq T\) is open on \(T\). For any point, \(p \in U\), as \(p \in U'\), there is a \(B_\alpha\) such that \(p \in B_\alpha \subseteq U'\) by the definition of basis. As \(p \in S\), \(p \in B_\alpha \cap S \subseteq U' \cap S = U\). So, for any open set, \(U \subseteq S\), and any point, \(p \in U\), there is a \(B_\alpha \cap S\) such that \(p \in B_\alpha \cap S \subseteq U\), which means that \(\{B_\alpha \cap S\}\) is a basis for \(S\) by the definition of basis.
References
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A description/proof of that open set complement of measure 0 subset is dense
Topics
About:
metric space
About:
measure
About:
map
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any open set on any metric space, the complement of any measure 0 subset with respect to the open set is dense on the open set.
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: Description
For any metric space, M, any open set, \(U \subseteq M\), and any measure 0 subset, \(S \subseteq U\), the complement, \(S^c = U \setminus S\), of \(S\) with respect to \(U\) is dense on \(U\).
2: Proof
Centered at any point, \(p \in U\), there is any small-enough open ball, \(B_{p-\epsilon} \subseteq U\). \(S^c \cap B_{p-\epsilon} = (U \setminus S) \cap B_{p-\epsilon} = U \cap B_{p-\epsilon} \setminus S \cap B_{p-\epsilon}\). \(m (S^c \cap B_{p-\epsilon}) = m (U \cap B_{p-\epsilon}) - m (S \cap B_{p-\epsilon}) = m (B_{p-\epsilon}) - 0\). As \(m (B_{p-\epsilon}) \gt 0\), \(m (S^c \cap B_{p-\epsilon}) > 0\), which means that \(S^c \cap B_{p-\epsilon}\) is not empty, which means that for any point, \(p \in U\), and any small-enough open ball, \(B_{p-\epsilon} \subseteq U\), there is a point of \(S^c\) inside the open ball, which means that \(S^c\) is dense.
References
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A description/proof of that from convex open set whose closure is bounded on Euclidean normed \(C^\infty\) manifold into equal or higher dimensional Euclidean normed \(C^\infty\) manifold polynomial map image of measure 0 subset is measure 0
Topics
About:
Euclidean normed \(C^\infty\) manifold
About:
measure
About:
map
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any polynomial map from any convex open set whose closure is bounded on any Euclidean normed \(C^\infty\) manifold, into any equal or higher dimensional Euclidean normd \(C^\infty\) manifold, the map image of any measure 0 subset is measure 0.
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: Description
For any Euclidean normed \(C^\infty\) manifolds, \(\mathbb{R}^n\) and \(\mathbb{R}^m\) where \(n \leq m\), any convex open set whose closure is bounded, \(U \subseteq \mathbb{R}^n\), any polynomial map, \(f: U \rightarrow \mathbb{R}^m\), and any measure \(0\) subset, \(S \subseteq U, m (S) = 0\) where \(m (\bullet)\) is the measure of the argument, the image of \(S\) under \(f\) is measure \(0\), which is \(m (f (S)) = 0\).
2: Proof
The polynomial map can be regarded to be defined on \(\mathbb{R}^n\), canonically extended from \(U\) to \(\mathbb{R}^n\). As \(f\) is \(C^1\), by the proposition that any \(C^1\) map from any open set on any Euclidean normed \(C^\infty\) manifold to any Euclidean normed \(C^\infty\) manifold satisfies the Lipschitz condition in any convex open set whose closure is bounded and contained in the original open set, \(f\) satisfies the Lipschitz condition on \(U\).
By the proposition that the image of any measure 0 subset under any Lipschitz condition satisfying map from any Euclidean normed topological space to any equal or higher dimensional Euclidean normed topological space is measure 0, \(m (f (S)) = 0\).
References
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A description/proof of that from Euclidean normed topological space into equal or higher dimensional Euclidean normed topological space Lipschitz condition satisfying map image of measure 0 subset is measure 0
Topics
About:
Euclidean normed topological space
About:
measure
About:
map
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the image of any measure 0 subset under any Lipschitz condition satisfying map from any Euclidean normed topological space to any equal or higher dimensional Euclidean normed topological space is measure 0.
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: Description
For any Euclidean normed topological spaces with the canonical metrics, \(\mathbb{R}^n\) and \(\mathbb{R}^m\) where \(n \leq m\), any open set, \(U \subseteq \mathbb{R}^n\), any Lipschitz condition satisfying map, \(f: U \rightarrow \mathbb{R}^m\), and any measure \(0\) subset, \(S \subseteq U\), the image of \(S\) under \(f\), \(f (S)\), is measure \(0\).
2: Proof
As \(f\) satisfies the Lipschitz condition, there is a constant, \(L\), such that for any points, \(p_1, p_2 \in U\), \(\Vert f (p_2) - f (p_1)\Vert \leq L \Vert p_2 - p_1 \Vert\). As \(S\) is measure \(0\), by the proposition that any area on any \(\mathbb{R}^n\) Euclidean metric space can be measured using only hypersquares, instead of hyperrectangles, for any real number, \(\epsilon \gt 0\), there are some countable disjoint half-open hypersquares, \(\{I_i\}\), \(S \subseteq \cup_i I_i, \sum_i m (I_i) \lt \epsilon\) where \(m (\bullet)\) is the measure of the argument. \(f (S) \subseteq f (\cup_i I_i) = \cup_i f (I_i)\) by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets, so, \(m (f (S)) \leq \sum _i m (f (I_i))\). For any points, \(p_1, p_2 \in I_i\), \(\Vert p_2 - p_1 \Vert \lt \sqrt{n {l_i}^2}\) where \(l_i\) is the side length of the hypersquare, so, \(\Vert f (p_2) - f (p_1) \Vert \leq L \Vert p_2 - p_1 \Vert \lt L \sqrt{n {l_i}^2}\), so, \(f (I_i)\) is contained in a \(2 L \sqrt{n {l_i}^2}\) side-length hypersquare very safely (meaning that there is a smaller possible hypersquare), so, \(m (f (I_i)) \lt (2 L \sqrt{n {l_i}^2})^m\). \(\sum_i m (f (I_i)) \lt \sum_i (2 L \sqrt{n {l_i}^2})^m = (2 L \sqrt{n})^m \sum_i {l_i}^m\), but as \(m (I_i) = {l_i}^n\), \(\sum_i m (f (I_i)) \lt (2 L \sqrt{n})^m \sum_i (m (I_i))^{m / n}\). For any small enough \(\epsilon\), \(m (I_i) \lt 1\), so, as \(m / n \geq 1\), \((m (I_i))^{m / n} \leq m (I_i)\), so, \(\sum_i (m (I_i))^{m / n} \lt \epsilon\), so, \(\sum_i m (f (I_i)) \lt (2 L \sqrt{n})^m \epsilon\). For any real number, \(\epsilon' \gt 0\), \(\epsilon\) can be chosen such that \(\epsilon' \gt (2 L \sqrt{n})^m \epsilon\), then, \(\sum_i m (f (I_i)) \lt \epsilon'\), so, \(m (f (S)) \lt \epsilon'\).
3: Note
When \(m \lt n\), the proposition does not hold. As a counterexample, let \(n = 3, m = 2\) where \(\mathbb{R}^2\) is the \(z = 0\) plane in \(\mathbb{R}^3\) and \(U = \mathbb{R}^3\) and \(f\) be the projection. \(f\) satisfies the Lipschitz condition with \(L = 1\). A square on the \(z = 0\) plane is measure \(0\) on \(\mathbb{R}^3\), and its image under \(f\) is the same square on \(\mathbb{R}^2\), whose measure is not \(0\) on \(\mathbb{R}^2\).
References
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A description/proof of that area of hyperrectangle can be approximated by area of covering finite number hypersquares to any precision
Topics
About:
Euclidean metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the area of any hyperrectangle can be approximated by the area of covering finite number hypersquares to any precision.
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: Description
The area, \(a\), of any hyperrectangle on any \(\mathbb{R}^n\) Euclidean metric space can be approximated by the area of some finite number of hypersquares, \(a_i\), that cover the hyperrectangle, to any precision, which is, for any real number \(\epsilon \gt 0\), \(\sum _i a_i - a \lt \epsilon\).
2: Proof
Denote the side lengths of the hyperrectangle as \(l_1, l_2, . . ., l_n\). For any real number, \(l_s \gt 0\), and each side index, \(1 \leq i \leq n\), there is the non-negative unique integer, \(k_i (l_s)\), and the unique real number, \(0 \leq l'_i (l_s) \lt l_s\), such that \(k_i (l_s) l_s = l'_i (l_s) + l_i\), which means that the \(l_i\) length is covered by \(k_i (l_s)\) times \(l_s\) with the excess, \(l'_i (l_s)\), where the unique existences are clear based on the meaning.
Now, \(l_s\) is the side length of the same-size hypersquares that cover the hyperrectangle, starting from a vertex of the hyperrectangle, \(k_i (l_s)\) times for each \(i\) direction, with the excess of \(l'_i (l_s)\) for the \(i\) direction.
The excess area is the error of the approximation, which is \(e (l_s) = \prod _i l_s k_i (l_s) - \prod _i l_i\). But by directly calculating the excess area, \(e (l_s) \lt \sum _i l'_i (l_s) \prod _{j \neq i} (l'_j (l_s) + l_j)\), adding some areas multiple times. Assuming \(l_s \leq 1\) (which we are free to do) , \(\sum _i l'_i (l_s) \prod _{j \neq i} (l'_j (l_s) + l_j) \lt \sum _i l_s \prod _ {j \neq i} (l_s + l_j) \lt l_s \sum _i \prod _ {j \neq i} (1 + l_j)\).
As \(L:= \sum _i \prod _ {j \neq i} (1 + l_j)\) depends only the configuration of the hyperrectangle, we choose \(l_s\) as \(min (1, \frac{\epsilon}{L})\), then, \(e (l_s) \lt l_s L \leq \epsilon\).
References
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A description/proof of that \(C^1\) map from open set on Euclidean normed \(C^\infty\) manifold to Euclidean normed \(C^\infty\) manifold locally satisfies Lipschitz condition
Topics
About:
Euclidean normed \(C^\infty\) manifold
About:
map
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any \(C^1\) map from any open set on any Euclidean normed \(C^\infty\) manifold to any Euclidean normed \(C^\infty\) manifold satisfies the Lipschitz condition in any convex open set whose closure is bounded and contained in the original open set.
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: Description
For any Euclidean normed \(C^\infty\) manifold, \(\mathbb{R}^{d_1}\), any open set, \(U \subset \mathbb{R}^{d_1}\), and any Euclidean normed \(C^\infty\) manifold, \(\mathbb{R}^{d_2}\), any \(C^1\) map, \(U \rightarrow \mathbb{R}^{d_2}\), satisfies the Lipschitz condition on any convex open set, \(U' \subseteq U\), whose closure is bounded and is contained in \(U\), which is, there is a constant, L, such that for any points, \(p_1, p_2 \in U'\), \(\Vert f (p_2) - f (p_1)\Vert \leq L \Vert p_2 - p_1\Vert\).
2: Proof
On \(U'\), by the mean value theorem for any differentiable function from any Euclidean normed \(C^\infty\) manifold to any Euclidean normed \(C^\infty\) manifold, \(\Vert f (p_2) - f (p_1)\Vert \leq \Vert Df (p_3)\Vert \Vert p_2 - p_1\Vert\) where \(p_3\) is a point on the line segment from \(p_1\) to \(p_2\) and \(\Vert Df (p_3)\Vert\) is the matrix norm (induced by vector norms) of the Jacobian. As each matrix element is continuous, it has the minimum and the maximum on the closure of \(U'\), because the bounded closet set is compact by the Heine-Borel theorem, and the image of any continuous map from any compact topological space to the \(\mathbb{R}\) Euclidean topological space has the minimum and the maximum, so, on \(U'\), each matrix element has a finite infimum and a finite supremum, so, the matrix norm has a finite supremum, L. So, \(\Vert f (p_2) - f (p_1)\Vert \leq L \Vert p_2 - p_1\Vert\).
References
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A description/proof of criteria for collection of open sets to be basis
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of some criteria for any collection of open sets to be a basis.
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: Description 1
For any topological space, \(T\), any collection of open sets, \(\{B_\alpha\}\), is a basis of \(T\), if and only if each open set on \(T\) is the union of some elements of the collection.
2: Proof 1
Suppose that each open set on \(T\) is the union of some elements of the collection. For any open set, \(U\), there is the union, \(\cup_{\alpha_1} B_{\alpha_1} = U\), and any \(B_{\alpha_1}\) satisfies \(B_{\alpha_1} \subseteq U\), so, the collection is a basis.
Suppose that the collection is a basis. Any open set, \(U\), is the union of some elements of the collection, because around any point, \(p \in U\), there is an open set, \(U_p \subseteq U\), by the local criterion for openness; there is an open set, \(B_\alpha \subseteq U_p\), in the basis, by the definition of basis; \(U\) is the union of such open sets in the basis.
3: Description 2
For any set, \(S\), and any collection of subsets, \(B = \{B_\alpha \subseteq S\}\), the collection of all the unions of elements of \(B\) (the empty set is included as the union of no element) constitutes a topology for \(S\), and \(B\) is a basis of the topological space, if and only if 1) \(S\) is the union of all the elements of \(B\), which is \(S = \cup B_\alpha\), and 2) for each sets, \(B_1, B_2 \in B\), and each point, \(p \in B_1 \cap B_2\), there is a set, \(B_3 \in B\), such that \(p \in B_3 \subseteq B_1 \cap B_2\).
4: Proof 2
Suppose that the conditions 1) and 2) are satisfied. In the collection of all the unions of elements of \(B\), \(S\) is included because of 1) and the empty set is included, and any possibly uncountably infinite union of elements of the collection is included in the collection, because it is a union of elements of \(B\), and any finite intersection of elements of the collection is included in the collection, because for any elements, \(\cup_\alpha B_\alpha\) and \(\cup_\beta B_\beta\), the intersection, \((\cup_\alpha B_\alpha) \cap (\cup_\beta B_\beta) = \cup_{\alpha, \beta} (B_\alpha \cap B_\beta)\), but because of 2), for any \(p \in B_\alpha \cap B_\beta\), there is a \(B_\gamma \in B\) such that \(p \in B_\gamma \subseteq B_\alpha \cap B_\beta\), so, \(B_\alpha \cap B_\beta = \cup B_\gamma\) as such \(p\)s cover \(B_\alpha \cap B_\beta\), so, the intersection is \(\cup B_\gamma\), which is in the collection. So, the collection is indeed a topology. \(B\) is a basis of the topology, because for any open set, which is a union of elements of \(B\), and any point in the open set, the point is in an element of \(B\), and the element satisfies the definition of \(B\)'s being a basis.
Suppose that the supposed topology is really a topology and \(B\) is a basis of the topological space. By Description 1, as \(S\) is open, \(S\) is the union of some elements of \(B\), but then, \(S\) is the union of all the elements of \(B\), because adding any other remaining element, which is a subset of \(S\), does not change the union. As \(B\) is a basis, \(B_1\) and \(B_2\) are open, and \(B_1 \cap B_2\) is open, and by the definition of basis, there is such a \(B_3\).
References
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A description/proof of that Euclidean topological space nested in Euclidean topological space is topological subspace
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any Euclidean topological space nested in any Euclidean topological space is a topological subspace of the nesting Euclidean topological space.
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: Description
For any Euclidean topological spaces, \(\mathbb{R}^{d_1}\) and \(\mathbb{R}^{d_2}\), such that \(\mathbb{R}^{d_2} \subseteq \mathbb{R}^{d_1}\) where inevitably \(d_2 \leq d_1\), where it does not mean that \(\mathbb{R}^{d_1} = \mathbb{R}^{d_2} \times \mathbb{R}^{d_1 - d_2}\), because \(\mathbb{R}^{d_2}\) may have be turned and/or translated against \(\mathbb{R}^{d_1}\), the topology of \(\mathbb{R}^{d_2}\) is the subspace topology of \(\mathbb{R}^{d_1}\).
2: Proof
It suffices to show that any open ball, \(B_1\), on \(\mathbb{R}^{d_1}\) is an open ball, \(B_2 = B_1 \cap \mathbb{R}^{d_2}\), or the empty set on \(\mathbb{R}^{d_2}\), and for any open ball, \(B_2\), on \(\mathbb{R}^{d_2}\), there is an open ball, \(B_1\), on \(\mathbb{R}^{d_1}\) such that \(B_2 = B_1 \cap \mathbb{R}^{d_2}\), because then, any open set, \(U_2\), on \(\mathbb{R}^{d_2}\) is a union of some open balls, \(\cup_{\alpha} B_{2_\alpha}\), on \(\mathbb{R}^{d_2}\), and each of the open balls will be open in the subspace topology, so, \(U_2\) will be open in the subspace topology, while any subset, \(S_2\), on \(\mathbb{R}^{d_2}\) that is open in the subspace topology is \(S_2 = U_1 \cap \mathbb{R}^{d_2}\) where \(U_1\) is open on \(\mathbb{R}^{d_1}\), but \(U_1\) is a union of some open balls, \(\cup_{\alpha} B_{1_\alpha}\), on \(\mathbb{R}^{d_1}\), and \(S_2 = (\cup_{\alpha} B_{1_\alpha}) \cap \mathbb{R}^{d_2} = \cup_{\alpha} (B_{1_\alpha} \cap \mathbb{R}^{d_2})\), but as \(B_{1_\alpha} \cap \mathbb{R}^{d_2}\) will be an open ball or empty set on \(\mathbb{R}^{d_2}\), \(S_2\) will be open on \(\mathbb{R}^{d_2}\).
Now, \(\mathbb{R}^{d_2}\) is the subset of \(\mathbb{R}^{d_1}\) where the \(d_2 + 1, . . ., d_1\) components are 0, which (the subset) is afterward turned around the origin and then translated. So, we can take the global chart on \(\mathbb{R}^{d_1}\) that is the standard chart turned around the origin and then translated in the same way with for \(\mathbb{R}^{d_2}\). In the new chart, \(\mathbb{R}^{d_2}\) is the subset of \(\mathbb{R}^{d_1}\) where the \(d_2 + 1, . . ., d_1\) components are 0, and any point on \(\mathbb{R}^{d_2}\) has the same \(1, . . ., d_2\) components with the \(\mathbb{R}^{d_2}\) standard chart and with the \(\mathbb{R}^{d_1}\) new chart. Let us always use the new chart from now on.
\(B_1\) is \(\{x \in \mathbb{R}^{d_1}| \sum_{i = 1, . . ., d_1} (x_i - p_i)^{2} \lt \epsilon^2\}\) where \(p\) is the center of the ball. \(B_1 \cap \mathbb{R}^{d_2}\) is \(\{x \in \mathbb{R}^{d_2}| \sum_{i = 1, . . ., d_2} (x_i - p_i)^{2} + \sum_{i = d_2 + 1, . . ., d_1} (p_i)^{2} \lt \epsilon^2\} = \{x \in \mathbb{R}^{d_2}| \sum_{i = 1, . . ., d_2} (x_i - p_i)^{2} \lt \epsilon^2 - \sum_{i = d_2 + 1, . . ., d_1} (p_i)^{2}\}\), which is an open ball or the empty set on \(\mathbb{R}^{d_2}\).
\(B_2\) is \(\{x \in \mathbb{R}^{d_2}| \sum_{i = 1, . . ., d_2} (x_i - p_i)^{2} \lt \epsilon^2\}\) where \(p\) is the center of the ball. There is a subset on \(\mathbb{R}^{d_1}\), \(\{x \in \mathbb{R}^{d_1}| \sum_{i = 1, . . ., d_2} (x_i - p_i)^{2} + \sum_{i = d_2 + 1, . . ., d_1} (x_i - 0)^{2} \lt \epsilon^2\}\), which is an open ball, named \(B_1\), on \(\mathbb{R}^{d_1}\) with the center \((p_1, . . ., p_{d_2}, 0, . . ., 0)\), and \(B_1 \cap \mathbb{R}^{d_2}\) is \(B_2\).
References
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A description/proof of that topological space is normal iff for closed set and its containing open set there is closed-set-containing open set whose ~
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About:
topological space
The table of contents of this article
Starting Context
Target Context
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The reader will have a description and a proof of the proposition that any topological space is normal if and only if for any closed set and its any containing open set, there is a closed-set-containing open set whose closure is contained in the former open set.
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: Description
Any topological space, \(T\), is normal if and only if for any closed set, \(C \subseteq T\), and any open set, \(U_1 \subseteq T\), such that \(C \subseteq U_1\), there is an open set, \(U_2 \subseteq T\), such that \(C \subseteq U_2 \subseteq \overline{U_2} \subseteq U_1\) where \(\overline{U_2}\) is the closure of \(U_2\).
2: Proof
Suppose that \(T\) is normal. Suppose that there are C and \(U_1\). \(C_2 := T \setminus U_1\) is closed and C and \(C_2\) are disjoint. There are some open sets, \(U_2, U_3 \subseteq T\), that are disjoint such that \(C \subseteq U_2\) and \(C_2 \subseteq U_3\). \(C_4 := T \setminus U_3\) is closed and \(C_2 = T \setminus U_1 \subseteq U_3 = T \setminus C_4\), so, \(C_4 \subseteq U_1\). As \(U_2\) and \(U_3\) are disjoint, \(U_2 \subseteq T \setminus U_3 = C_4\). So, \(U_2 \subseteq C_4 \subseteq U_1\), but as \(\overline{U_2}\) is the intersection of all the closed sets that contain \(U_2\), \(U_2 \subseteq \overline{U_2} \subseteq C_4\). So, \(C \subseteq U_2 \subseteq \overline{U_2} \subseteq U_1\).
Suppose that for C and \(U_1\), there is \(U_2\) such that \(C \subseteq U_2 \subseteq \overline{U_2} \subseteq U_1\). For any disjoint closed sets, \(C_1, C_2 \subseteq T\), \(U_1 := T \setminus C_2\) is open and \(C_1 \subseteq U_1\). So, there is an open set, \(U_2 \subseteq T\), such that \(C_1 \subseteq U_2 \subseteq \overline{U_2} \subseteq U_1\). As \(\overline{U_2}\) is closed, \(U_3 := T \setminus \overline{U_2}\) is open and \(T \setminus U_3 = \overline{U_2} \subseteq U_1 = T \setminus C_2\), which implies \(C_2 \subseteq U_3\). As \(U_3 = T \setminus \overline{U_2} \subseteq T \setminus U_2\), \(U_2\) and \(U_3\) are disjoint.
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