- #1
brydustin
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Is an open normed subspace Y (subset of X) primarily defined as a set {y in X : Norm(y) < r}? Where r is some real (positive) number.
I know the open ball definitions and such... but it seems like this definition is saying, an open normed space, is essentially an open ball which satisfies the criteria of a normed vector space.
I don't like this definition, as given a vector, say g, whose norm is arbitrarily close to r, we can always double the vector, as 2*g. The norm is greater than r so it can't be included in Y, yet it should be included in Y as a subspace must be closed under scalar multiplication (and vector addition).
However, in "Linear Functional Analysis" by Bryan Rynne and Martin Youngson, we are asked first to prove
a) If there is n > 0 such that {y in X: norm(y) < n} subset of Y, show that nx/(2*norm(x)) is in Y. This is easy to show.
b)If Y is open, then Y=X.
The problem I have is with the solution in the back of the book...
Let x be in X. As Y is open there exists n > 0 such that {y in X: norm(y) < n} subset of Y. Hence, nx/(2*norm(x)) by part (a). As a scalar multiple of a vector in Y is also in Y, we have that x = (2norm(x)/n )(nx/(2*norm(x))) is in Y... therefore, X is a subset of Y. And Y is a subset of X (by definition)... therefore X=Y.
The part I don't understand is why we need Y to be open to have n > 0 s.t. {y in X: norm(y) < n} contained in Y. After that, everything is obvious... Y ISNT defined as that "open ball" as I say at the beginning:
"Is an open normed subspace Y (subset of X) primarily defined as a set {y in X : Norm(y) < r}? Where r is some real (positive) number."
... but I don't see how to make sense of it all. Thanks in advance
I know the open ball definitions and such... but it seems like this definition is saying, an open normed space, is essentially an open ball which satisfies the criteria of a normed vector space.
I don't like this definition, as given a vector, say g, whose norm is arbitrarily close to r, we can always double the vector, as 2*g. The norm is greater than r so it can't be included in Y, yet it should be included in Y as a subspace must be closed under scalar multiplication (and vector addition).
However, in "Linear Functional Analysis" by Bryan Rynne and Martin Youngson, we are asked first to prove
a) If there is n > 0 such that {y in X: norm(y) < n} subset of Y, show that nx/(2*norm(x)) is in Y. This is easy to show.
b)If Y is open, then Y=X.
The problem I have is with the solution in the back of the book...
Let x be in X. As Y is open there exists n > 0 such that {y in X: norm(y) < n} subset of Y. Hence, nx/(2*norm(x)) by part (a). As a scalar multiple of a vector in Y is also in Y, we have that x = (2norm(x)/n )(nx/(2*norm(x))) is in Y... therefore, X is a subset of Y. And Y is a subset of X (by definition)... therefore X=Y.
The part I don't understand is why we need Y to be open to have n > 0 s.t. {y in X: norm(y) < n} contained in Y. After that, everything is obvious... Y ISNT defined as that "open ball" as I say at the beginning:
"Is an open normed subspace Y (subset of X) primarily defined as a set {y in X : Norm(y) < r}? Where r is some real (positive) number."
... but I don't see how to make sense of it all. Thanks in advance