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Given an infinite cardinal b such that $b^2=b$ I need to find a cardinal $x$ such that:

$$(x^b = 2^b)\quad \land \quad (\forall c.(c>x)\to (c^b>2^b))$$

Furthermore, I need to show there is only one such $x$.

I think the answer is $ x = 2^b$ but have trouble proving the second part:$$(c>2^b) \to (c^b >2^b) $$

Any thoughts?

  • 2
    What exactly is $b$? – 2017-01-01
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    $c>2^b \rightarrow c^b \geq c> 2^b $ – 2017-01-01
  • 0
    I don't know what $b$ is, but presumably if there's only one $x$ such that $x^b=2^b$ and (statement) then that $x$ is $2$. – 2017-01-01
  • 0
    b and c are infinite cardinals – 2017-01-01
  • 0
    I edited the question, forgot a part of the question – 2017-01-01
  • 0
    The question is missing nothing. – 2017-01-02

1 Answers 1

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Certainly $x=2^b$ satisfies the given conditions: $x^b=\left(2^b\right)^b=2^{b^2}=2^b=x$, and if $c>x=2^b$, then $c^b\ge c>2^b$. To show uniqueness, suppose that $y$ is an infinite cardinal such that $y^b=2^b$, and $c^b>2^b$ whenever $c>y$. Then on the one hand we must have $y\le x$, since $y^b=2^b$, but on the other hand we must have $x\le y$, since $x^b=2^b$, so $x=y$.