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I have to answer the following question:

Let $\{x_n\}_{n\ge 1}$ be a Schauder Basis. Is $\{x_{\Pi(n)}\}_{n\ge 1}$, where $\Pi(n)$ bijection of $\mathbb N$ onto $\mathbb N$, also a Schauder Basis?

It seems to me that it is not true, but I can't show that formally. Can you help me? I'm completely green in this topic.

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    It's not, necessarily. See [this](http://math.stackexchange.com/questions/1560262/rearrangement-of-schauder-basis) for an example. You may also want to google "unconditional Schauder basis". – 2017-01-19

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A Schauder Basis $\{x_n\}_{n\ge 1}$ of the Banach space $X$ is called an unconditional basis if for each $x \in X$ the series

$x= \sum_{n}\xi_n x_n$

is unconditionally convergent. The space $C ( [ 0 , 1 ] )$ has a Schauder basis but no unconditional Schauder basis .

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    thanks everybody! It starts to be clear but I must think about it more – 2017-01-19