Proving Same Cardinality of F(\mathbb{Q},\mathbb{R}) and \mathbb{R}

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SUMMARY

The discussion centers on proving that the cardinality of the set of functions from the rational numbers \(\mathbb{Q}\) to the real numbers \(\mathbb{R}\), denoted as \(F(\mathbb{Q},\mathbb{R})\), is equal to the cardinality of \(\mathbb{R}\). Participants emphasize the necessity of establishing a bijection between these two sets to demonstrate their equivalence in cardinality. The cardinal number of \(F(\mathbb{Q},\mathbb{R})\) is identified as \(2^{\aleph_0}\), which matches the cardinality of \(\mathbb{R}\).

PREREQUISITES
  • Understanding of cardinality and bijections in set theory
  • Familiarity with the concepts of \(\mathbb{Q}\) (rational numbers) and \(\mathbb{R}\) (real numbers)
  • Knowledge of functions and mappings in mathematics
  • Basic grasp of set notation and terminology
NEXT STEPS
  • Research the concept of cardinality in set theory
  • Learn about bijections and their role in proving set equivalences
  • Explore the properties of the continuum hypothesis related to \(\mathbb{R}\)
  • Study the implications of \(F(X,Y)\) for different sets \(X\) and \(Y\)
USEFUL FOR

Mathematicians, students studying set theory, and anyone interested in the properties of cardinal numbers and functions between sets.

burak100
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Homework Statement





F(\mathbb{Q},\mathbb{R}) is the set of maps from \mathbb{Q} to \mathbb{R}. Then show that F(\mathbb{Q},\mathbb{R}) and \mathbb{R} have same potency (cardinal number?)..



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The Attempt at a Solution



I am no tsure but I think I need to find bijection map between these sets, but how?
 
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What is the cardinal number of F(\mathbb{Q},\mathbb{R})??
 

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