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Prove that ℝ has no subspaces except ℝ and {0}.
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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?
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?
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}?
Is (-2,2) not a sunspace because all the scalers are of the natural numbers?
True
No 3 is not. The sunspace (-2,2) does not contain all of the real numbers of R. For example 3.
It is not a sunspace because it is not closed under multiplication.
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.