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Can someone give me some hints for proving that $∀\varepsilon >0, x+\varepsilon > t\Rightarrow x\geq t$? I really don’t know how to start with. Thanks

Christopher S.
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  • Let $x+\epsilon>t$ for all $\epsilon >0$ but assume (for contradiction) that $x<t$. – AloneAndConfused Sep 28 '17 at 22:44
  • try this question: https://math.stackexchange.com/questions/438477/discrete-mathematics-x-leq-y-epsilon-implies-x-leq-y –  Sep 28 '17 at 23:29

1 Answers1

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Hint:

Suppose $x < t$, then $t-x > 0$.

Siong Thye Goh
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