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Open mappings

  1. Feb 15, 2012 #1

    jgens

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    1. The problem statement, all variables and given/known data

    Let [itex](X,||\cdot||)[/itex] be a normed vector space and suppose that [itex]Y[/itex] is a closed vector subspace of [itex]X[/itex]. Show that the map [itex]||x||_1=\inf_{y \in Y}||x-y||[/itex] defines a pseudonorm on [itex]X[/itex]. Let [itex](X/Y,||\cdot||_1)[/itex] denote the normed vector space induced by [itex]||\cdot||_1[/itex] and prove that the canonical projection [itex]\phi:X \rightarrow X/Y[/itex] is an open mapping.

    2. Relevant equations

    All vector spaces are over [itex]\mathbb{R}[/itex].

    3. The attempt at a solution

    So far I have been able to show that [itex]||\cdot||_1[/itex] is a pseudonorm, but I am having difficulty showing the canonical projection is an open mapping. Obviously we need to take [itex]U[/itex] open in [itex]X[/itex] and then take [itex][x] \in \phi(U)[/itex]. From here we need to construct an open neighborhood around [itex][x][/itex] which is contained in [itex]\phi(U)[/itex] but I am having difficulty doing this. Any help?
     
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  3. Feb 15, 2012 #2

    AKG

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    What does the fact that [itex]x\in U[/itex] with [itex]U[/itex] open tell you? Based on what that gets you, you should be able to think of a candidate for the desired open neighbourhood containing [itex][x][/itex] and contained in [itex]\phi(U)[/itex].
     
  4. Feb 15, 2012 #3

    jgens

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    Well I know that there exists [itex]0 < \varepsilon[/itex] such that [itex]B(x,\varepsilon) \subseteq U[/itex]. But the problem I then have is that [itex]||\cdot||_1 \leq ||\cdot||[/itex] so I am having trouble finding a criterion with [itex]||x-y||_1 < \delta[/itex] implies [itex]||x-y|| < \varepsilon[/itex].
     
  5. Feb 16, 2012 #4

    micromass

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    We wish to prove that [itex]B_1(\varphi(x),\varepsilon)\subseteq \varphi(B(x,\varepsilon))[/itex]. We can take x=0 if we want.

    So take [itex]z\in B_1(0,\varepsilon)[/itex], write out what that means.
     
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