If $a+\epsilon > b$ for each $\epsilon >0$, can we conclude that $a>b$?
Please help me to clarify the above. Thanks in advance.
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$\begingroup$ Well, $a \geq b.$ $\endgroup$– Will JagyCommented Sep 29, 2013 at 4:10
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8$\begingroup$ Be careful: "for every $\epsilon > 0$, if $a+\epsilon > b$, then $a > b$" is a different statement than "if, for every $\epsilon > 0$, $a + \epsilon > b$, then $a > b$." One of them is a lot closer to being true than the other. $\endgroup$– MicahCommented Sep 29, 2013 at 4:25
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1$\begingroup$ Related: Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$?. Some of the posts linked there might be of interest, too. $\endgroup$– Martin SleziakCommented May 3, 2017 at 12:30
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$\begingroup$ Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? $\endgroup$– user1110341Commented May 27, 2023 at 15:19
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1 Answer
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No, you cannot: for every $\epsilon>0$ it’s true that $1+\epsilon>1$, but it’s not true that $1>1$. What you can conclude is that $a\ge b$. To see this, suppose that $a<b$, and let $\epsilon=b-a$; then $\epsilon>0$, but $a+\epsilon=a+(b-a)=b\not>b$, contradicting the hypothesis. Thus, it must be the case that $a\ge b$.
answered Sep 29, 2013 at 4:13
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$\begingroup$ Suppose $a = 5$ and $b = 10$. Now what? $\endgroup$ Commented Sep 29, 2013 at 5:17
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4$\begingroup$ @Don: Then obviously it’s not the case that $a+\epsilon>b$ for each $\epsilon>0$, so the example is irrelevant. $\endgroup$ Commented Sep 29, 2013 at 5:19
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1$\begingroup$ Yeah Don with the example your using you go beyond the hpyothesis imagaine in ur example if u chose epsilon to = 15 then it would work since 5 + 15 = 20 > 10 ie 5+ϵ > 10, but here we had to use a particular epsilon choice. The whole point is we wanted it to work for ANY epsilon that a person want to choose, no matter how small or how big as long as it is greater than zero. to techniquely u wnat ur choice to be as small as possible her to make sure its holds for any epislon, if it works for the smallest number(dont exist so we use ϵ) it obvisly works for larger numbers $\endgroup$– user114392Commented Dec 8, 2013 at 18:56
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1$\begingroup$ @alu: It’s a proof by contradiction. We are given that $a+\epsilon>a$ for each positive $\epsilon$. To get a contradiction, we suppose that $a<b$. Then for $\epsilon=b-a$, which is positive, we have a+\epsilon=b$. This contradicts what we were given. Thus, the supposition that $a<b$ must be false, since it is what led us to the contradiction, and therefore $a\ge b$. $\endgroup$ Commented Jan 15, 2021 at 18:57
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1$\begingroup$ @alu: It was supposed to say: To get a contradiction, we suppose that $a<b$. then for $\epsilon=b-a$, which is positive, we have $a+\epsilon=b$. This contradicts what we were given. That is, we were told that $a+\epsilon$ is always greater than $b$ when $\epsilon>0$, and now we’ve found a counterexample. $\endgroup$ Commented Jan 18, 2021 at 7:11