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MDolphins
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Prove that ℝ has no subspaces except ℝ and {0}.
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MDolphins said:Would you use the property that a W is a subspace of V if and only if W is closed under vector addition and scalar mulitplication in V?
MDolphins said:Well, if we were to look at a subspace that is not in ℝ, it would not be closed under the same addition or multiplication that is in ℝ. And additionally, from the theorem "if a subset S of a vector space V does not contain the zero vector 0 of V, then S is not a subspace of V". From this, the 0 vector of the ℝ is the set 0 itself. THerefore, the subspace must be ℝ and {0}.
DOes this seem right?
MDolphins said:Well, the closure property holds for v, because since v is a part of the reals, any real number times v will still be apart of the reals. Therefore, the only subspace of the reals are the reals and {0}?
MDolphins said:Is (-2,2) not a sunspace because all the scalers are of the natural numbers?
MDolphins said:True
MDolphins said:No 3 is not. The sunspace (-2,2) does not contain all of the real numbers of R. For example 3.
MDolphins said:It is not a sunspace because it is not closed under multiplication.
MDolphins said:So going back to the proof, the real numbers and 0 are the only subspaces of R bc they seethe only sets that hold under. Scalar multiplication.
ℝ, also known as the set of real numbers, is the set of all numbers that can be represented on a number line. It includes all rational and irrational numbers.
A subspace is a subset of a vector space that is closed under addition and scalar multiplication. In simpler terms, it is a smaller set that still follows the rules of a larger set.
Proving this statement helps us understand the structure of ℝ and its subspaces. It also helps us identify and differentiate between different types of vector spaces and their properties.
This can be proved by showing that any other subset of ℝ does not meet the criteria of a subspace, specifically it is not closed under addition and scalar multiplication.
The proof that ℝ has no subspaces except ℝ and {0} tells us that ℝ is a simple and complete vector space with no proper subspaces. This has implications in fields such as linear algebra and functional analysis.