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Functional Analysis: Open normed subspace

  1. Feb 14, 2012 #1
    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
     
  2. jcsd
  3. Feb 14, 2012 #2

    micromass

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    No, this is never done. A subspace is supposed to be closed under addition and multiplication. An open ball is clearly not. So it would make very little sense to define an open subspace like that.

    What is your definition of an open set?? You can define a set to be open if for each point there is an n such that [itex]\{y\in X~\vert~\|y\|<n\}\subseteq Y[/itex]. That's one possible definition.
    All depends on how you define an open set.
     
  4. Feb 14, 2012 #3
    I don't see how this is different than your definition below......

    I don't see how it "depends", a set is either open or not. The solution says, AS Y is open there exists n such that [itex]\{y\in X~\vert~\|y\|<n\}\subseteq Y[/itex]. But I don't see why it must be the case.
     
  5. Feb 14, 2012 #4

    morphism

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    "Y is open" means that around each point of Y there is an open ball (of some radius) contained entirely within Y. This is NOT the same as Y being an open ball. For example the upper half-plane is an open subset of R^2 (but of course isn't a vector subspace). Is it an open ball?

    The solution is merely taking an open ball around 0. (Since Y is a subspace, it contains 0. And because Y is open, it contains an open ball around 0.)
     
  6. Feb 14, 2012 #5
    Your examples are clear.... but I don't see why we need Y to be open to have an open ball of radius n . Could we let Y be closed and still fit a ball of radius n inside of it, then continue with the "proof". .... I don't see how this proves anything.
    Sorry
     
  7. Feb 14, 2012 #6

    morphism

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    What exactly are you asking?

    Are you asking "why is an open subspace Y of a normed space X necessarily equal to X"? This is because an open subspace will contain a ball around 0, hence vectors "pointing in all directions", and therefore (by scaling) it must be the entire space. More rigorously, given an x in X, we can find some nonzero scalar k such that kx is in a given open ball around 0 (k could potentially be very small). But if we do this to get kx inside the open ball about 0 contained in Y, then we'd in particular get that kx is in Y, and therefore that x=1/k(kx) is in Y as well.

    Or are you really asking "why does Y need to be open to be have an open ball of radius n"? The answer is of course it doesn't! For example, in R^2, the closed upper half-plane {(x,y) | y >= 0} isn't open (why not?) but certainly contains many open balls. (By contrast, look at the set Z^2 of points in R^2 with integer coordinates. This set doesn't contain a single open ball.)

    However, if a set is open, then it HAS to contain an open ball around each of its points. (Note that the closed half-plane above doesn't contain an open ball around any point on the x-axis.) In particular, if you have an open subspace Y, then it must contain a ball around each of its points, i.e. given y in Y, we can find an r>0 such that the ball [itex]\{ x \in X \mid \|x-y\|< r \}[/itex] is fully contained in Y. In particular, we can find such a ball centered at y=0...
     
  8. Feb 14, 2012 #7

    Thanks... that makes sense ^_^
     
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