386: Writer '.uno:InsertPagebreak'

<The previous article in this series | The table of contents of this series | The next article in this series>

A UNO dispatch command that inserts the page break at the current cursor

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Topics

About: UNO (Universal Network Objects)
About: LibreOffice
About: Apache OpenOffice
About: Writer

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: The Specifications

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Starting Context

• The reader has knowledge of what 'UNO dispatch command' is and how to call one (if the URL and the arguments of the command are clear).

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Target Context

• The reader will know the specifications of the UNO dispatch command.

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Orientation

There are some articles that explain how to execute any UNO dispatch commands and get the whole available information from the execution, in Java, in C++, in C#, in Python, and in LibreOffice or Apache OpenOffice Basic.

There are the list for the application foundations, the list for Writer, and the list for Calc of the UNO dispatch commands listed so far in this series.

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Main Body

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1: The Specifications

URL: .uno:InsertPagebreak

Description: This command inserts the page break at the current view cursor.

Arguments (the types are UNO datum types):

nothing

The related information (the value of com.sun.star.frame.FeatureStateEvent.State) (the types are UNO datum types):

Type	Value
N/A	void

The result information (the value of com.sun.star.frame.DispatchResultEvent.Result) (the type is a UNO datum type):

Type	Value
N/A	void

<The previous article in this series | The table of contents of this series | The next article in this series>

385: Writer '.uno:RemoveDirectCharFormats'

<The previous article in this series | The table of contents of this series | The next article in this series>

A UNO dispatch command that removes the direct characters formats at the current view cursor

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Topics

About: UNO (Universal Network Objects)
About: LibreOffice
About: Apache OpenOffice
About: Writer

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: The Specifications

---

Starting Context

• The reader has knowledge of what 'UNO dispatch command' is and how to call one (if the URL and the arguments of the command are clear).

---

Target Context

• The reader will know the specifications of the UNO dispatch command.

---

Orientation

There are some articles that explain how to execute any UNO dispatch commands and get the whole available information from the execution, in Java, in C++, in C#, in Python, and in LibreOffice or Apache OpenOffice Basic.

There are the list for the application foundations, the list for Writer, and the list for Calc of the UNO dispatch commands listed so far in this series.

---

Main Body

---

1: The Specifications

URL: .uno:RemoveDirectCharFormats

Description: This command removes the direct characters formats at the current view cursor, when nothing is selected.

Arguments (the types are UNO datum types):

nothing

The related information (the value of com.sun.star.frame.FeatureStateEvent.State) (the types are UNO datum types):

Type	Value
N/A	void

The result information (the value of com.sun.star.frame.DispatchResultEvent.Result) (the type is a UNO datum type):

Type	Value
N/A	void

<The previous article in this series | The table of contents of this series | The next article in this series>

379: Subnet of Net with Directed Index Set

<The previous article in this series | The table of contents of this series | The next article in this series>

A definition of subnet of net with directed index set

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition

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Starting Context

• The reader knows a definition of net with directed index set.
• The reader knows a definition of final map between directed sets.

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Target Context

• The reader will have a definition of subnet of net with directed index set.

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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 directed set, $S_1$, any topological space, $T$, and any net with directed index set, $f_1:S_1\to T$, any net with directed index set, $f_1\circ f_2:S_2\to T$ where $S_2$ is any directed set and $f_2:S_2\to S_1$ is any final map between directed sets

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

384: Closed Set on Closed Topological Subspace Is Closed on Base Space

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A description/proof of that closed set on closed topological subspace is closed on base space

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of topological space.
• The reader knows a definition of subspace topology.
• The reader knows a definition of closed set.

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Target Context

• The reader will have a description and a proof of the proposition that any closed set on any closed topological subspace is closed on the base 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 topological space, $T_1$, and any closed topological subspace, $T_2\subseteq T_1$, of $T_1$, any closed set, $C\subseteq T_2$, on $T_2$, is closed on $T_1$.

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2: Proof

As $C$ is closed on $T_2$, $T_2\setminus C$ is open on $T_2$. For any point, $p\in T_2\setminus C$, there is an open neighborhood, $U_p\subseteq T_2\setminus C$, of $p$ on $T_2$. By the definition of subspace topology, $U_p=U_p^{\prime }\cap T_2$ where $U_p^{\prime }\subseteq T_1$ is open on $T_1$. But $U_p^{\prime }=U_p^{\prime }\cap (T_2\cup (T_1\setminus T_2))=(U_p^{\prime }\cap T_2)\cup (U_p^{\prime }\cap (T_1\setminus T_2))\subseteq (T_2\setminus C)\cup (T_1\setminus T_2)$, because $U_p^{\prime }\cap T_2\subseteq T_2\setminus C$ and $(U_p^{\prime }\cap (T_1\setminus T_2))\subseteq T_1\setminus T_2$. As $(T_2\setminus C)\cup (T_1\setminus T_2)=T_1\setminus C$, $U_p^{\prime }\subseteq T_1\setminus C$, which means that around any point on $T_2\setminus C$, there is an open neighborhood on $T_1$ that is contained in $T_1\setminus C$. As $T_2$ is closed on $T_1$, $T_1\setminus T_2$ is open on $T_1$. For any point, $p\in T_1\setminus T_2$, there is an open neighborhood, $U_p\subseteq T_1\setminus T_2\subseteq T_1\setminus C$, of $p$ on $T_1$, which means that around any point on $T_1\setminus T_2$, there is an open neighborhood on $T_1$ that is contained in $T_1\setminus C$. As $T_1\setminus C=(T_2\setminus C)\cup (T_1\setminus T_2)$, around any point on $T_1\setminus C$, there is an open neighborhood on $T_1$ that is contained in $T_1\setminus C$, which means that $T_1\setminus C$ is open on $T_1$, so, $C$ is closed on $T_1$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

383: Open Set on Open Topological Subspace Is Open on Base Space

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A description/proof of that open set on open topological subspace is open on base space

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of topological space.
• The reader knows a definition of subspace topology.

---

Target Context

• The reader will have a description and a proof of the proposition that any open set on any open topological subspace is open on the base 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 topological space, $T_1$, and any open topological subspace, $T_2\subseteq T_1$, of $T_1$, any open set, $U\subseteq T_2$, on $T_2$, is open on $T_1$.

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2: Proof

By the definition of subspace topology, $U=U^{\prime }\cap T_2$ where $U^{\prime }\subseteq T_1$ is open on $T_1$. As $T_2$ is open on $T_1$, $U$ is open on $T_1$ as an intersection of finite open sets.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

382: Continuous Embedding

<The previous article in this series | The table of contents of this series | The next article in this series>

A definition of continuous embedding

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition
• 2: Note

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Starting Context

• The reader knows a definition of continuous map.
• The reader knows a definition of injection.
• The reader knows a definition of homeomorphism.

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Target Context

• The reader will have a definition of continuous embedding.

---

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 spaces, $T_1,T_2$, any continuous map, $f:T_1\to T_2$, such that $f$ is an injection and the codomain restriction, $f^{\prime }:T_1\to f(T_1)$ is a homeomorphism

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2: Note

'Continuous embedding' and '$C^{\mathrm{\infty }}$ embedding' are different, while any $C^{\mathrm{\infty }}$ embedding is a continuous embedding, a continuous embedding is not necessarily a $C^{\mathrm{\infty }}$ embedding.

Usually just 'embedding' is used as it is usually obvious which: non-$C^{\mathrm{\infty }}$ map cannot be a $C^{\mathrm{\infty }}$ embedding and (just) embedding-ness of a $C^{\mathrm{\infty }}$ map is customarily understood to be $C^{\mathrm{\infty }}$ embedding-ness.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

380: Accumulation Value of Net with Directed Index Set Is Convergence of Subnet

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that accumulation value of net with directed index set is convergence of subnet

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Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of convergence of net with directed index set.
• The reader knows a definition of subnet of net with directed index set.
• The reader knows a definition of accumulation value of net with directed index set.

---

Target Context

• The reader will have a description and a proof of the proposition that any accumulation value of any net with directed index set is the convergence of a subnet.

---

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 directed set, $S_1$, any topological space, $T$, any net with directed index set, $f_1:S_1\to T$, and any accumulation value of $f_1$, $p\in T$, there is a subnet of $f_1$, $f_1\circ f_2:S_2\to T$ where $S_2$ is a directed set and $f_2:S_2\to S_1$, such that $p$ is a convergence of $f_1\circ f_2$.

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2: Proof

Let us define $S_2$ as $\{(\alpha ,N_p)|\alpha \in S_1,N_p\text{ is any neighborhood of }p\text{ such that }{f}_1(\alpha )\in N_p\}$ with the relation such that $({\alpha }_1,N_{p-1})\le ({\alpha }_2,N_{p-2})$ if and only if ${\alpha }_1\le {\alpha }_2$ and $N_{p-2}\subseteq N_{p-1}$, where there is at least 1 such $N_p$ for each $\alpha$, because there is a neighborhood of $p$ and there is a neighborhood of $f_1(\alpha )$ and the union of the 2 neighborhoods is a one ($N_p$ does not need to be connected). Note that all the possible pairs are included in $S_2$.

$S_2$ is indeed a directed set, because 1) $({\alpha }_1,N_{p-1})\le ({\alpha }_1,N_{p-1})$; 2) if $({\alpha }_1,N_{p-1})\le ({\alpha }_2,N_{p-2})$ and $({\alpha }_2,N_{p-2})\le ({\alpha }_3,N_{p-3})$, $({\alpha }_1,N_{p-1})\le ({\alpha }_3,N_{p-3})$; 3) for any $({\alpha }_1,N_{p-1})$ and $({\alpha }_2,N_{p-2})$, there is a $({\alpha }_3,N_{p-3})$ such that $({\alpha }_1,N_{p-1})\le ({\alpha }_3,N_{p-3})$ and $({\alpha }_2,N_{p-2})\le ({\alpha }_3,N_{p-3})$, because by the definition of directed set, there is an ${\alpha }_4$ such that ${\alpha }_1\le {\alpha }_4$ and ${\alpha }_2\le {\alpha }_4$ and as $p$ is an accumulation value, there is an ${\alpha }_3$ such that ${\alpha }_4\le {\alpha }_3$ and $f_1({\alpha }_3)\in N_{p-1}\cap N_{p-2}:=N_{p-3}$, then ${\alpha }_1\le {\alpha }_3$ and ${\alpha }_2\le {\alpha }_3$ and $N_{p-3}\subseteq N_{p-1}$ and $N_{p-3}\subseteq N_{p-2}$.

Let us define $f_2$ as $(\alpha ,N_p)\mapsto \alpha$, which is final, because for any $\alpha$, $\alpha \le f_2((\alpha ,N_p))=\alpha$.

$p$ is a convergence of $f_1\circ f_2$, because for any neighborhood of $p$, $N_{p-0}$, there is a $({\alpha }_0,N_{p-0})$ because $p$ is an accumulation value of $f_1$, and for every $({\alpha }_0,N_{p-0})\le (\alpha ,N_p)$, $f_1\circ f_2((\alpha ,N_p))=f_1(\alpha )\in N_{p-0}$, because $f_1(\alpha )\in N_p\subseteq N_{p-0}$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

378: Final Map Between Directed Sets

<The previous article in this series | The table of contents of this series | The next article in this series>

A definition of final map between directed sets

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition

---

Starting Context

• The reader knows a definition of directed set.

---

Target Context

• The reader will have a definition of final map between directed 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 directed sets, $S_1,S_2$, any map, $f:S_1\to S_2$, such that for any $p_2\in S_2$, there is a $p_1\in S_1$ such that $p_2\le f(p_1)$

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

377: Accumulation Value of Net with Directed Index Set

<The previous article in this series | The table of contents of this series | The next article in this series>

A definition of accumulation value of net with directed index set

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Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition
• 2: Note

---

Starting Context

• The reader knows a definition of net with directed index set.

---

Target Context

• The reader will have a definition of accumulation value of net with directed index 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: Definition

For any directed set, $S$, any topological space, $T$, and any net with directed index set, $f:S\to T$, any point, $p\in T$, such that for any neighborhood, $N_p\subseteq T$, of $p$ and any index $i_0\in S$, there is an index, $i\in S$, such that $i_0\le i$ and $f(i)\in N_p$

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2: Note

A net with directed index set can have multiple accumulation values.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

376: For Hausdorff Topological Space, Net with Directed Index Set Can Have Only 1 Convergence

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that for Hausdorff topological space, net with directed index set can have only 1 convergence

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Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of Hausdorff topological space.
• The reader knows a definition of convergence of net with directed index set.

---

Target Context

• The reader will have a description and a proof of the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most.

---

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 directed set, $S$, any Hausdorff topological space, $T$, and any net with directed index set, $f:S\to T$, $f$ can have only 1 convergence at most.

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2: Proof

Suppose $f$ had 2 convergences, $p_1,p_2\in T$. For any neighborhood of $p_j$ where $j=1\text{ or }2$, $N_j$, there would be an index, $i_j\in S$, such that $f(i)\in N_j$ for every $i\in S,i_j\le i$. As $T$ is Hausdorff, let us take $N_1$ and $N_2$ as disjoint. By the definition of directed set, there would be an index, $i_3\in S$, such that $i_1\le i_3$ and $i_2\le i_3$. For every $i_3\le i$, $f(i)\in N_1$ and $f(i)\in N_2$, which is a contradiction, as $N_1$ and $N_2$ would be disjoint.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

375: For Metric Space, 1 Point Subset Is Closed

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A description/proof of that for metric space, 1 point subset is closed

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Topics

About: metric space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of metric space.
• The reader knows a definition of closed subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any metric space, any 1 point subset is closed.

---

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 1 point subset, $\{p\}\subseteq M$, is closed.

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2: Proof

Think of $M\setminus \{p\}$ and any point, $p^{\prime }\in M\setminus \{p\}$. As $p\ne p^{\prime }$, by the definition of distance between 2 points, $d(p,p^{\prime })>0$. So, for any $0<ϵ<d(p,p^{\prime })$, the open ball around $p^{\prime }$, $B_{p^{\prime }-ϵ}$, does not contain $p$, so, $B_{p^{\prime }-ϵ}\subseteq M\setminus \{p\}$. So, $M\setminus \{p\}$ is open, so, $\{p\}$ is closed.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

371: For Metric Space, Difference of Distances of 2 Points from Subset Is Equal to or Less Than Distance Between Points

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that for metric space, difference of distances of 2 points from subset is equal to or less than distance between points

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Topics

About: metric space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of metric space.
• The reader knows a definition of distance between 2 points on metric space.
• The reader knows a definition of distance between point and subset on metric space.

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Target Context

• The reader will have a description and a proof of the proposition that for any metric space, any subset, and any 2 points, the difference of the distances of the points from the subset is equal to or less than the distance between the 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 metric space, $M$, any subset, $S\subseteq M$, and any points, $p_1,p_2\in M$, $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$ where $d(\bullet ,\bullet )$ denotes the distance between the arguments.

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2: Proof

Suppose that $p_1,p_2\notin S$. Take any point, $p_2^{\prime }\in S$. By the definition of distance between 2 points, $d(p_1,p_2^{\prime })\le d(p_1,p_2)+d(p_2,p_2^{\prime })$. By the definition of distance between point and subset, $d(p_1,S)\le d(p_1,p_2^{\prime })$ and $d(p_2,p_2^{\prime })=d(p_2,S)+ϵ$ where $ϵ\ge 0$ can be made any small positive. So, $d(p_1,S)\le d(p_1,p_2)+d(p_2,S)+ϵ$. By taking the limit by $ϵ\to 0$, $d(p_1,S)\le d(p_1,p_2)+d(p_2,S)$, so, $d(p_1,S)-d(p_2,S)\le d(p_1,p_2)$. By the symmetricalness, also $d(p_2,S)-d(p_1,S)\le d(p_1,p_2)$, so $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$.

Suppose that $p_1\notin S$ and $p_2\in S$. By the definition of distance between point and subset, $d(p_1,S)\le d(p_1,p_2)$. As $d(p_2,S)=0$, $d(p_1,S)-d(p_2,S)\le d(p_1,p_2)$. As the left side is non-negative, $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$.

Suppose that $p_1\in S$ and $p_2\notin S$. By the symmetricalness, $|d(p_2,S)-d(p_1,S)|\le d(p_2,p_1)$, which implies $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$.

Suppose that $p_1,p_2\in S$. As $d(p_1,S)=d(p_2,S)=0$, $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

370: Identity Map with Domain and Codomain Having Different Topologies Is Continuous iff Domain Is Finer than Codomain

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that identity map with domain and codomain having different topologies is continuous iff domain is finer than codomain

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of topological space.
• The reader knows a definition of finer and coarser topologies.
• The reader knows a definition of continuous map.

---

Target Context

• The reader will have a description and a proof of the proposition that any identity map with domain and codomain having the same set but having different topologies is continuous if and only if the domain topology is finer than codomain's.

---

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 spaces, $T_1,T_2$, that have the same set but have different topologies and the identity map, $f:T_1\to T_2$, $f$ is continuous if and only if the $T_1$ topology is finer than the $T_2$ topology.

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2: Proof

Suppose that the $T_1$ topology is finer than the $T_2$ topology. Take any open set, $U\subseteq T_2$. The preimage $f^{-1}(U)=U$, is open on $T_1$, because the $T_1$ topology is finer, so, $f$ is continuous.

Suppose that $f$ is continuous. Take any open set, $U\subseteq T_2$. The preimage $f^{-1}(U)=U$, is open on $T_1$, because $f$ is continuous, so, the $T_1$ topology is finer.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

369: For Regular Topological Space, Neighborhood of Point Contains Closed Neighborhood

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that for regular topological space, neighborhood of point contains closed neighborhood

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Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of regular topological space.

---

Target Context

• The reader will have a description and a proof of the proposition that for any regular topological space, any neighborhood of any point contains a closed neighborhood.

---

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 regular topological space, $T$, any point, $p\in T$, and any neighborhood of $p$, $p\in N_p\subseteq T$, $N_p$ contains a closed neighborhood.

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2: Proof

There is an open set, $U_p\subseteq N_p,p\in U_p$. $T\setminus U_p$ is closed and does not contain $p$. So, by the definition of regular topological space, there are some disjoint neighborhoods, $N_1,N_2$, such that $p\in N_1\subseteq T$ and $T\setminus U_p\subseteq N_2\subseteq T$, $N_1\cap N_2=\mathrm{\varnothing }$ where we take $N_1$ and $N_2$ as open neighborhoods, which is always possible (see the Note of the definition article). In fact, $\overline{N_1}\cap N_2=\mathrm{\varnothing }$ where $\overline{N_1}$ is the closure of $N_1$, because $\overline{N_1}=N_1\cup ac(N_1)$ where $ac(N_1)$ is the set of all the accumulation points of $N_1$, but any accumulation point of $N_1$ does not belong to $N_2$, because for any point, $p^{\prime }\in N_2$, as $N_2$ is open, there is an open set, $U_{p^{\prime }}$, $p^{\prime }\in U_{p^{\prime }}\subseteq N_2$, which does not contain any point of $N_1$, so, $p^{\prime }$ is not any accumulation point of $N_1$. So, $\overline{N_1}\subseteq T\setminus N_2\subseteq T\setminus (T\setminus U_p)=U_p$, because $T\setminus U_p\subseteq N_2$. So, $\overline{N_1}$ is a closed neighborhood contained in $U_p$, which is contained in $N_p$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

368: Map That Is Anywhere Locally Constant on Connected Topological Space Is Globally Constant

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A description/proof of that map that is anywhere locally constant on connected topological space is globally constant

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of connected topological space.
• The reader admits the proposition that any pair of elements of any open cover of any connected topological space is finite-open-sets-sequence-connected via some elements of the open cover.

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Target Context

• The reader will have a description and a proof of the proposition that any map that is anywhere locally constant on any connected topological space is globally constant.

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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.

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Main Body

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1: Description

For any connected topological space, $T$, and any map, $f:T\to S$ where $S$ is any set, such that around each point, $p\in T$, there is an open set, $U_p\subseteq T,p\in U_p$, such that $f$ is constant on $U_p$, $f$ is constant globally.

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2: Proof

The set of $U_p$s for all the points constitutes an open cover, $S_c=\{U_p\}$.

Choose any point, $p_0\in T$, and there is the open set, $U_{p_0}\in S_c,p_0\in U_{p_0}$, on which $f$ takes the constant image, $f(p_0)$. For any other point, $p\in T$, there is the open set, $U_p\in S_c,p\in U_p$, on which $f$ takes the constant image, $f(p)$.

By the proposition that any pair of elements of any open cover of any connected topological space is finite-open-sets-sequence-connected via some elements of the open cover, there is a sequence of elements of $S_c$ starting at $U_{p_0}$ and ending at $U_p$ where each adjoining pair of elements of the sequence shares a point. Obviously, via those shared points, $f$ takes the same image, $f(p_0)$, on all the elements of the sequence. So, $f(p)=f(p_0)$.

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3: Note

Although the proposition may seem obvious, we have to exactly prove the existence of such a sequence.

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References

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367: Regular Topological Space

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A definition of regular topological space

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition
• 2: Note

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Starting Context

• The reader knows a definition of topological space.
• The reader knows a definition of closed set.
• The reader knows a definition of neighborhood of point.
• The reader knows a definition of neighborhood of subset.

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Target Context

• The reader will have a definition of regular topological space.

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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.

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Main Body

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1: Definition

Any topological space, $T$, such that each 1 point subset is closed, and for any point, $p\in T$, and any closed set that does not contain $p$, $C\subseteq T,p\notin C$, there are disjoint neighborhoods, $p\in N_1\subseteq T$ and $C\subseteq N_2\subseteq T$, $N_1\cap N_2=\mathrm{\varnothing }$

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2: Note

In fact, $N_1$ and $N_2$ can be taken as open neighborhoods, because we can take open neighborhoods contained in $N_1$ and $N_2$, which preserves the disjointness.

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References

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366: Pair of Elements of Open Cover of Connected Topological Space Is Finite-Open-Sets-Sequence-Connected Via Cover Elements

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A description/proof of that pair of elements of open cover of connected topological space is finite-open-sets-sequence-connected via cover elements

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

---

Starting Context

• The reader knows a definition of connected topological space.
• The reader knows a definition of finite-open-sets-sequence-connected pair of open sets.

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Target Context

• The reader will have a description and a proof of the proposition that any pair of elements of any open cover of any connected topological space is finite-open-sets-sequence-connected via some elements of the open cover.

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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.

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Main Body

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1: Description

For any connected topological space, $T$, and any open cover, $S_c=\{U_{\alpha }\},{\cup }_{\alpha }{U}_{\alpha }=T$, any pair of elements, $U_1,U_2\in S_c$, are finite-open-sets-sequence-connected via some elements of $S_c$.

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2: Proof

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, $({\cup }_{\beta }{U}_{\beta })\cap ({\cup }_{\gamma }{U}_{\gamma })\ne \mathrm{\varnothing }$ 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 $p\in U_{\beta }$ 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 $U_1$ via some elements of $S_c$. So, $S_e$ equals $S_c$.

So, $U_2\in S_e$, and $U_1$ and $U_2$ are finite-open-sets-sequence-connected via some elements of $S_c$.

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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.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

365: Pair of Open Sets of Connected Topological Space Is Finite-Open-Sets-Sequence-Connected

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A description/proof of that pair of open sets of connected topological space is finite-open-sets-sequence-connected

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

---

Starting Context

• The reader knows a definition of connected topological space.
• The reader knows a definition of finite-open-sets-sequence-connected pair of open sets.

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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

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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.

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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, $({\cup }_{\beta }{U}_{\beta })\cap ({\cup }_{\gamma }{U}_{\gamma })\ne \mathrm{\varnothing }$ 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.

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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.

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References

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