932: When Union of Simplicial Complexes Is Simplicial Complex, Underlying Space of Union Is Union of Underlying Spaces of Constituents

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description/proof of that when union of simplicial complexes is simplicial complex, underlying space of union is union of underlying spaces of constituents

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Note
• 4: Proof

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Starting Context

• The reader knows a definition of simplicial complex.

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Target Context

• The reader will have a description and a proof of the proposition that when the union of some simplicial complexes is a simplicial complex, the underlying space of the union is the union of the underlying spaces of the constituent complexes.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$V_1$: $\in \{\text{ the real vectors spaces }\}$
$V_2$: $\in \{\text{ the real vectors spaces }\}$
$C_1$: $\in \{\text{ the simplicial complexes on }{V}_1\}$
$C_2$: $\in \{\text{ the simplicial complexes on }{V}_2\}$
//

Statements:
(
$V_1\in \{\text{ the subspaces of }{V}_2\}\vee V_2\in \{\text{ the subspaces of }{V}_1\}$
$\wedge$
$C_1\cap C_2\in \{\text{ the simplicial complexes on }{V}_1\cup V_2\}$
)
$⟹$
$|C_1\cup C_2|=|C_1|\cup |C_2|$
//

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2: Natural Language Description

For any real vectors spaces, $V_1,V_2$, such that $V_1$ is a vectors subspace of $V_2$ or $V_2$ is a vectors subspace of $V_1$, and any simplicial complexes, $C_1$, on $V_1$ and $C_2$, on $V_2$, when $C_1\cup C_2$ is a simplicial complex on $V_1\cup V_2$, $|C_1\cup C_2|=|C_1|\cup |C_2|$.

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3: Note

As has been shown in the proposition that the union of some simplicial complexes is not necessarily a simplicial complex, $C_1\cup C_2$ is not necessarily a simplicial complex. This is about when $C_1\cup C_2$ is so.

Compare with the proposition that the intersection of any 2 simplicial complexes is a simplicial complex, and the underlying space of the intersection is contained in but not necessarily equal to the intersection of the underlying spaces of the constituent simplicial complexes.

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4: Proof

Whole Strategy: Step 1: see that $|C_1\cup C_2|\subseteq |C_1|\cup |C_2|$; Step 2: see that $|C_1|\cup |C_2|\subseteq |C_1\cup C_2|$.

Step 1:

Let $p\in |C_1\cup C_2|$ be any. There is an $S\in C_1\cup C_2$ such that $p\in S$. $S\in C_1$ or $S\in C_2$. $p\in |C_1|$ or $p\in |C_2|$, so, $p\in |C_1|\cup |C_2|$.

Step 2:

Let $p\in |C_1|\cup |C_2|$ be any. $p\in |C_1|$ or $p\in |C_2|$. There is an $S_1\in C_1$ such that $p\in S_1$ or an $S_2\in C_2$ such that $p\in S_2$. If the former holds, as $S_1\in C_1\cup C_2$, $p\in |C_1\cup C_2|$; otherwise, likewise, $p\in |C_1\cup C_2|$, so, $p\in |C_1\cup C_2|$ anyway.

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References

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