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I want to show $\sqrt{x}$ is uniformly continuous on $[0, \infty)$. I know it is uniformly continuous on $[0,1]$ and I can show it's uniformly cts. on $[1, \infty)$, so if I choose $x=1/2$ and $y\in [1,\infty)$, is it uniformly continuous on $[0,\infty)$?

Babai
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user282271
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    What you wrote is a series of statements with no connection between them. A mathematical proof this is not. Also, you cannot just "choose" $x$. What you need to prove is the statement: "For all $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x,y$, if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon$." You did not prove this statement. – 5xum Oct 20 '15 at 19:52
  • I'm asking about the idea of the proof, not the details – user282271 Oct 20 '15 at 19:54
  • I understand that. But in order to discuss what you already did, you need to write that in a coherent manner, not by simply throwing some symbols around. – 5xum Oct 20 '15 at 20:04

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Use the fact that $$|\sqrt{|x|} -\sqrt{|y|} |\leqslant \sqrt{|x-y|}$$ for any $x,y\in\mathbb{R} .$