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For example,in " $a\leq b+\epsilon$ if $\forall\epsilon>0$, then $a\leq b$ ",it's impossible to find a case where $a=b+\epsilon$ for very $\epsilon>0$.However,people are still using $\leq$ even though $<$ is enough to describe that. Why is that?

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    $\begingroup$ There are contexts where $<$ is more appropriate and contexts where $\le$ is more appropriate. $\endgroup$ – Mauro ALLEGRANZA May 8 at 8:19
  • $\begingroup$ Please avoid all caps words. Thank you. $\endgroup$ – Wojowu May 8 at 8:27
  • $\begingroup$ In cases such that $<$ and $\le$ do not make a difference, there is no reason to prefer one over the other. $\endgroup$ – Yves Daoust May 8 at 8:29
  • $\begingroup$ In your very example, observe that we cannot use < in conclusion 'a≤b'. So there you have one example where you still need to use ≤. $\endgroup$ – Wojowu May 8 at 8:32
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It is often a lot harder to show $<$ and a lot easier to show $\leq$. For example, if you have a sequence $(a_n)_{n \in \mathbb{N}}$ that is bounded above by a constant $C$, so $a_n \leq C$ for all $n$. If further the sequence converges, then also $$\lim\limits_{n \to \infty} a_n \leq C.$$

On the other hand, even if $a_n < C$ for all $n$, you can't conclude that $$\lim\limits_{n \to \infty} a_n < C.$$

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There may be many reasons, one could be that it is the more general statement. Another, that this way it is easier to remember, you don't need to remember which of the instances is $\le$ and which is $<$.

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The statement you ask about doesn't make sense as written, but presumably it is meant to say

If $a\leq b+\epsilon$ for every $\epsilon>0$ then $a\leq b$

In this case you can't replace the second $\leq$ by $<$ (even if you also change the first), since we could have $a=b$.

You could replace the first $\leq$ by $<$, but the original version is a stronger statement, so preferable. (For any given $\epsilon$, $a\leq b+\epsilon$ is a weaker condition than $a<b+\epsilon$, and making the conditions weaker gives a stronger statement.)

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  • $\begingroup$ exactly what i want to know,thx. $\endgroup$ – Max K May 8 at 10:31

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