f : X --> R I proved: $\displaystyle \exists M\geq 0, \forall \epsilon>0, \forall x\in X, M-\epsilon < |f(x)|< M+\epsilon$
Last edited by szpengchao; Oct 19th 2008 at 07:05 AM.
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Originally Posted by szpengchao f : X --> R I proved: $\displaystyle \exists M\geq 0, \forall \epsilon>0, \forall x\in X, M-\epsilon < |f(x)|< M+\epsilon$ I think you actually probably proved $\displaystyle M - \epsilon < f(x) < M + \epsilon$ for $\displaystyle x\in X$. Yes, this proves that $\displaystyle f$ is bounded on $\displaystyle X$.
Originally Posted by szpengchao f : X --> R I proved: $\displaystyle \exists M\geq 0, \forall \epsilon>0, \forall x\in X, M-\epsilon < |f(x)|< M+\epsilon$ If you can show that $\displaystyle \exists M\geq 0, \forall \epsilon>0, \forall x\in X, |f(x)|< M+\epsilon$ then it follows that $\displaystyle \forall x\in X, |f(x)|\le M$ or $\displaystyle f$ is bounded on $\displaystyle X$.
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