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When attempting to prove that a number is the infimum of a set, the proof normally follows as follows:

1)show set is bounded below by $t$.
2)assume $s$ is also a lower bound, but $s>t$
3)show $s$ is in the set therefore obtaining a contradiction.

However in this question: show that $\inf \{2 + \frac 2 n \mid n \in \Bbb N\} = 2$ using the definition of the infimum, I am confused by this part of the proof and do not understand it:

continuing from 2) let $s>2$ be a lower bound. Then for some $\varepsilon > 0$ we have $t = 2 + \varepsilon$ and for all $n \in \Bbb N$ we have $t = 2 + \varepsilon < 2 + \frac 3 n$.

But if $\varepsilon = 5$ then does the inequality $t = 2 + \varepsilon < 2 + \frac 3 n$ still hold (as $n=1$ leads to $7<5$)?

Many thanks.

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I'm assuming you're laying out the steps to show that $t$ is the infimum of a set. You say:

1)show set is bounded below by $t$.

2)assume $s$ is also a lower bound, but $s>t$

3)show $s$ is in the set therefore obtaining a contradiction.

Step 3 needs some alteration. It is not a contradiction to have $s$ in the set. It is a contradiction to have something in the set that is less than $s$, because this implies $s$ cannot be an upper bound.

Now concerning your proof, You haven't said what $t$ is, but you say $t=2+\epsilon$. Did you mean $s=2+\epsilon$? Also, the next step (for all $n \in \mathbb{N}$, $2+\epsilon<2+\frac{3}{n}$) is not true, as you pointed out.

Starting from the same point, if we assume $s>2$ is an lower bound for the set, then $s=2+\epsilon$ for some $\epsilon>0$. Based on my comments about step 3, we should look for an element of the set that is less than $s=2+\epsilon$. Every element in the set is of the form $2+\frac{2}{n}$ for some $n \in \mathbb{N}$, so we want to find $n$ such that $2+\frac{2}{n}<2+\epsilon$. This is accomplished if $\frac{2}{n}<\epsilon$. Can you find $n$ so that this inequality holds?

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