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I'm studying for Real Analysis and I can't understand the application of Lemma for covering sets numbers.

Decide whether cover $\zeta$ of the space X has a covering number. If so, find the set of all covering numbers for $\zeta$. $X=(-\infty,1)$ and $\zeta=\{{G_n\}}$ where $G_n=(n^2,1)$ for all $n$ belonging to natural numbers

If anyone could please explain the mechanics behind finding this (I know this means that $\inf\{{\delta(x): x {\rm\ belongs\ to\ }X\}}=0$ , but how do I find that?) I would be very grateful.

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Welcome to math.SE. $\LaTeX$ is freely available on the web. But you don't need any software to use $\TeX$ here. If you write your mathematics in $\TeX$ markup, the site will compile it automatically. – Brett Frankel Apr 19 '12 at 3:21
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A Lebesgue number for your open cover is a number $\delta>0$ such that any interval of $X$ of diameter less than $\delta$ is contained in some member of the given open cover.

So, if $X$ has a Lebesgue number, then given any interval in $X$ that is "small enough", there is some element of the cover containing it. Now, if you look at the sets in your open cover, you'll note that they do not contain the "left hand bit" of $(0,1)$. In fact, any interval of the form $(0,\delta)$ is contained in no element of the cover. From this we might intuit that $X$ has no Lebesgue number.

And, in fact, $X$ has no Lebesgue number. To rigorously show this, you might argue as follows:

You need to show that no Lebesgue number $\delta$ exists. Towards this end, let $\delta$ be any positive number. Now you need to find an interval in $X$ of diameter less than $\delta$ that is not contained in any element of the open cover. To do this, just consider the open interval centered at $(0,\delta/2)$.

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Thanks; my professor explained this in term of sup{r: x-r,x+r subset of Gn} and I understand that better. – Blodgett Apr 19 '12 at 3:56
OP has made an edit to the problem that makes parts of this answer incongruent. – Gerry Myerson May 10 '12 at 23:25

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