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The full question reads: Suppose that $a$ is a number that has the property that for every $n \in \mathbb{N}$, $a \leq 1/n$. Prove $a \leq 0$.

Is there anyway to show this using Archimedean Property, or is it something related to the Completeness Axiom? The problem using the Archimedean Property is that I get up to $a< \epsilon$ but from there I am not able to conclude anything about whether $a \leq 0$ because $\epsilon > 0$.

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HINT: Archimedean property. –  Arturo Magidin Sep 17 '11 at 23:48
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For any $\varepsilon>0$ exists $n\in\mathbb{N}$ such that $0<\frac{1}{n}<\varepsilon$ –  Frank Murphy Sep 18 '11 at 4:01
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People are really coming out of the woodwork to answer this! –  The Chaz 2.0 Sep 18 '11 at 12:50
    
It is nice that $\mathbb{R}$ is first-countable, isn't it? :) –  Mark Sep 18 '11 at 13:06
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"Anyway" is a perfectly respectable word, but it should not be used as a synonym of "any way" (two words). But in the last few years I've seen a number of cases of people doing this. –  Michael Hardy Sep 18 '11 at 14:39

6 Answers 6

up vote 8 down vote accepted

In other words, prove that $a\not>0$. Assume $a>0$, what does that tell you about some $n$?

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I'll not give the complete answer to this (on purpose).

But here is some intuition: if you have a positive number, what can you say about that positive number as compared to the numbers $1/n$ for each n?

Start with something concrete, like 0.1 for example. It can't have the property you mention... why not?

The same reason works for all positive numbers.

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If $a \leq 1/n$ for all $n \in \mathbb{N}$, then $a \le \inf \{ 1/n : n \in \mathbb{N} \}=0$. Now, to prove that the inf is $0$, you need the Archimedean Property. (The inf being 0 is actually equivalent to the Archimedean Property.)

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Note also that this uses the Completeness Axiom. In fact, the Completeness Axiom implies the Archimedean Property. –  lhf Sep 18 '11 at 15:30

Suppose that $a>0$. Then by the Archimedean Property, there exists a natural number $n$ so that $1/n<a$. But this is a contradiction. Then $a\leq0$ if $a\leq 1/n$ for all $n$.

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Assume $a>0$,then there exists some $\epsilon$ such that $0<\epsilon<a$.

Let $N=[\epsilon]+1$,then $\frac{1}{N}<\epsilon<a$, which contradicts to that for every $n∈N, a≤1/n$.

So, $a\leq0$.[Q.E.D]

Above is implied in @Daniel's and @lhf's answers.

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$n$ is a natural number, therefore it can be $1,2,3, \ldots$. On the other hand, $1/n$ is a continuous function for every real $x$ distinct to zero.

Applying the following limit $$ \mathop {\lim }\limits_{n \to \infty } a \le \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0 $$

$$ a \le 0 $$

$$Q.E.D.$$

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$1/n$ is not continuous. –  ThomasMcLeod Sep 18 '11 at 1:01
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Whatever that means. –  Did Sep 18 '11 at 8:01
    
it is continuous for every x distinct to zero, as stated above. –  Daniel Sep 18 '11 at 14:28
    
The problem states $n \in \mathbb{N}$. –  ThomasMcLeod Sep 20 '11 at 4:28

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