Which separable Banach space has no Schauder basis? [closed]

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Can you state concisely the essential idea behind Enflo's example of a separable Banach space with no Schauder basis?







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closed as off-topic by user21820, user91500, Gibbs, amWhy, Shailesh Aug 30 at 13:03


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    Can you state concisely the essential idea behind Enflo's example of a separable Banach space with no Schauder basis?







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    closed as off-topic by user21820, user91500, Gibbs, amWhy, Shailesh Aug 30 at 13:03


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, user91500, Gibbs, amWhy, Shailesh
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      Can you state concisely the essential idea behind Enflo's example of a separable Banach space with no Schauder basis?







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      Can you state concisely the essential idea behind Enflo's example of a separable Banach space with no Schauder basis?









      share|cite|improve this question











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      asked Aug 26 at 12:55









      H. Tomasz Grzybowski

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      closed as off-topic by user21820, user91500, Gibbs, amWhy, Shailesh Aug 30 at 13:03


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, user91500, Gibbs, amWhy, Shailesh
      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by user21820, user91500, Gibbs, amWhy, Shailesh Aug 30 at 13:03


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, user91500, Gibbs, amWhy, Shailesh
      If this question can be reworded to fit the rules in the help center, please edit the question.




















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          1. Every Banach space with a Schauder basis has the bounded approximation property, (b.a.p) that is, there exists a net of finite-rank operators $S_n$ such that $S_n$ tends to the identity operator in the strong operator topology.

          2. Alexander Grothendieck proved that every Banach space with the b.a.p also has the approximation property -- that is, every compact operator can be approximated by finite rank operators in the strong operator topology.

          3. Enflo constructs a separable Banach space which fails to have a slightly weaker property than the b.a.p, and in particular fails to have the b.a.p, which by combining Grothendieck's result with (1) implies that it fails to have a Schauder basis.

          It does not get much more concise than this.






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            1 Answer
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            1 Answer
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            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes








            up vote
            0
            down vote













            1. Every Banach space with a Schauder basis has the bounded approximation property, (b.a.p) that is, there exists a net of finite-rank operators $S_n$ such that $S_n$ tends to the identity operator in the strong operator topology.

            2. Alexander Grothendieck proved that every Banach space with the b.a.p also has the approximation property -- that is, every compact operator can be approximated by finite rank operators in the strong operator topology.

            3. Enflo constructs a separable Banach space which fails to have a slightly weaker property than the b.a.p, and in particular fails to have the b.a.p, which by combining Grothendieck's result with (1) implies that it fails to have a Schauder basis.

            It does not get much more concise than this.






            share|cite|improve this answer
























              up vote
              0
              down vote













              1. Every Banach space with a Schauder basis has the bounded approximation property, (b.a.p) that is, there exists a net of finite-rank operators $S_n$ such that $S_n$ tends to the identity operator in the strong operator topology.

              2. Alexander Grothendieck proved that every Banach space with the b.a.p also has the approximation property -- that is, every compact operator can be approximated by finite rank operators in the strong operator topology.

              3. Enflo constructs a separable Banach space which fails to have a slightly weaker property than the b.a.p, and in particular fails to have the b.a.p, which by combining Grothendieck's result with (1) implies that it fails to have a Schauder basis.

              It does not get much more concise than this.






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                1. Every Banach space with a Schauder basis has the bounded approximation property, (b.a.p) that is, there exists a net of finite-rank operators $S_n$ such that $S_n$ tends to the identity operator in the strong operator topology.

                2. Alexander Grothendieck proved that every Banach space with the b.a.p also has the approximation property -- that is, every compact operator can be approximated by finite rank operators in the strong operator topology.

                3. Enflo constructs a separable Banach space which fails to have a slightly weaker property than the b.a.p, and in particular fails to have the b.a.p, which by combining Grothendieck's result with (1) implies that it fails to have a Schauder basis.

                It does not get much more concise than this.






                share|cite|improve this answer












                1. Every Banach space with a Schauder basis has the bounded approximation property, (b.a.p) that is, there exists a net of finite-rank operators $S_n$ such that $S_n$ tends to the identity operator in the strong operator topology.

                2. Alexander Grothendieck proved that every Banach space with the b.a.p also has the approximation property -- that is, every compact operator can be approximated by finite rank operators in the strong operator topology.

                3. Enflo constructs a separable Banach space which fails to have a slightly weaker property than the b.a.p, and in particular fails to have the b.a.p, which by combining Grothendieck's result with (1) implies that it fails to have a Schauder basis.

                It does not get much more concise than this.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 26 at 13:51









                uniquesolution

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                8,271823












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