MHB Infinite dimentional subspace need not be closed

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The discussion centers on the properties of the subspace Y, defined as the span of polynomials in the space X = C[0,1]. One participant asserts that Y is not closed in X, while another argues that it is closed because it includes all linear combinations of polynomials, which remain within X. The disagreement highlights a misunderstanding about the closure of polynomial spans in the context of continuous functions. The conversation emphasizes the importance of clarity in mathematical definitions and properties. Ultimately, the closure of Y in X remains a contentious point among the participants.
nwl
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Let X=C[O,1] and Y=span($X_{0},X_{1},···$), where $X_{j}={t}^{i}$, so that Y is the set of all polynomials. Y is not closed in X.
 
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nwl said:
Let X=C[O,1] and Y=span(x\_{0},x\_{1},···), where X\_{j}={t}^{i}, so that Y is the set of all polynomials. Y is not closed in X.
Are you asking a question here or are you simply informing us? We can't help you if we don't know what you are asking about!

-Dan
 


I would have to disagree with your statement that Y is not closed in X. Y is indeed closed in X, as it is the span of all polynomials in X. This means that any linear combination of polynomials in X will also be a polynomial in X, and thus Y is a subset of X. Since Y is a subset of X and contains all polynomials in X, it is closed in X. Therefore, Y is closed in X.
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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