Talk:Uniform boundedness principle: Difference between revisions
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MathKnight (talk | contribs) The orginal non-Latex text of the proof |
AndrewKepert (talk | contribs) note on revert |
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:Since ''X'' is a [[Baire space]], one of the ''X<sub>n</sub>'' has an interior point, giving some δ > 0 such that ||''x''|| < δ ⇒ ''x'' ∈ ''X<sub>n</sub>''. |
:Since ''X'' is a [[Baire space]], one of the ''X<sub>n</sub>'' has an interior point, giving some δ > 0 such that ||''x''|| < δ ⇒ ''x'' ∈ ''X<sub>n</sub>''. |
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:Hence for all ''T'' ∈ ''F'', ||''T''|| < ''n''/δ, so that ''n''/δ is a uniform bound for the set ''F''. |
:Hence for all ''T'' ∈ ''F'', ||''T''|| < ''n''/δ, so that ''n''/δ is a uniform bound for the set ''F''. |
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::I am reverting to the non-latex version, because the latex version adds nothing. (It has actually subtracted something: a rather important "implies" sign.) See [[Wikipedia:How_to_write_a_Wikipedia_article_on_Mathematics]]. [[User:AndrewKepert|Andrew Kepert]] 22:46, 14 Mar 2005 (UTC) |
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Revision as of 22:46, 14 March 2005
What the hell?! --JensMueller
The orginal non-Latex text of the proof
- For n = 1,2,3, ... let Xn = { x : ||T(x)|| ≤ n (∀ T ∈ F) } . By hypothesis, the union of all the Xn is X.
- Since X is a Baire space, one of the Xn has an interior point, giving some
δ > 0 such that ||x|| <δ ⇒ x ∈ Xn. - Hence for all T ∈ F, ||T|| < n/
δ , so that n/δ is a uniform bound for the set F.
- I am reverting to the non-latex version, because the latex version adds nothing. (It has actually subtracted something: a rather important "implies" sign.) See Wikipedia:How_to_write_a_Wikipedia_article_on_Mathematics. Andrew Kepert 22:46, 14 Mar 2005 (UTC)