Showing when a subset is an equality

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Homework Help Overview

The discussion revolves around the conditions under which the subset relation \( A \subseteq \mathcal{P} \bigcup A \) holds as an equality. Participants are exploring the implications of set theory concepts, particularly focusing on the cardinalities of sets and their power sets.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • One participant attempts to analyze the cardinality of \( A \) and its implications for \( \mathcal{P} \bigcup A \), questioning the conditions for equality. Others raise clarifications about the definitions of the union and power set, indicating potential misunderstandings.

Discussion Status

The discussion is ongoing, with participants questioning definitions and clarifying terms. There is no explicit consensus yet, but the dialogue is exploring important distinctions in set theory.

Contextual Notes

There appears to be some confusion regarding the notation used, particularly between \( \mathcal{P} \bigcup A \) and \( \bigcup \mathcal{P} A \). Additionally, the nature of \( A \) as a single object is under scrutiny, which may affect the interpretation of the problem.

Mr Davis 97
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Homework Statement


In general, ##A \subseteq \mathcal{P} \bigcup A##. Under what conditions does equality hold?

Homework Equations

The Attempt at a Solution


I can't seem to figure this out. If ##A## has ##n## elements, then clearly ##| \bigcup A | \ge n##, which would mean that ##| \mathcal{P} \bigcup A | \ge 2^n##, right? In this case the cardinalities never seem to be the same, in which case equality can never hold.
 
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## \mathcal{P} \bigcup A ## and ##\bigcup \mathcal{P} A ## are different things.
 
mfb said:
## \mathcal{P} \bigcup A ## and ##\bigcup \mathcal{P} A ## are different things.
Oops. In both cases I meany ##\mathcal{P} \bigcup A##
 
What is ##\bigcup A##? The union of what, if A is a single object?
 

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