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evinda
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Hello! ![Cool :cool: :cool:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I want to show that if $x_n',x_n'' \in A$ with $x_n'-x_n'' \to 0 \Rightarrow f(x_n'')-f(x_n'') \to 0$,then $f$ is uniformly continuous on $A$.
We suppose that $f$ is not uniformly continuous o $A$.
So, $ \exists \epsilon>0$ such that $\forall \delta>0$ and $ \forall y_n',y_n'' \in A$ with $|y_n'-y_n''|<\delta$ $\Rightarrow$ $|f(y_n')-f(y_n'')| \geq \epsilon$.
Is it right so far? And how can I continue? (Thinking)
I want to show that if $x_n',x_n'' \in A$ with $x_n'-x_n'' \to 0 \Rightarrow f(x_n'')-f(x_n'') \to 0$,then $f$ is uniformly continuous on $A$.
We suppose that $f$ is not uniformly continuous o $A$.
So, $ \exists \epsilon>0$ such that $\forall \delta>0$ and $ \forall y_n',y_n'' \in A$ with $|y_n'-y_n''|<\delta$ $\Rightarrow$ $|f(y_n')-f(y_n'')| \geq \epsilon$.
Is it right so far? And how can I continue? (Thinking)