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Let $A= (0, 1]$ and let $B={(\frac{n+2}{2^n}, 2^{1/n}): n \in \Bbb N)}$

Please help I am having trouble on where to start because I'm still having a hard time on understanding open covers and subcovers.

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To see $B$ is a cover of $(0,1]$: we must pick an $x\in (0,1]$ and show it is contained in some $(\frac{n+2}{2^n},2^{1/n})$. Notice that $2^{1/n}>1$ so we must only find $n$ such that $\frac{n+2}{2^n}

To see no finite subcover exists: Define $b_n$ as $(\frac{n+2}{2^n},2^{1/n})$. Suppose that a finite subcover $C$ exist, pick an integer $n$ such that $\frac{n+2}{2^n}\leq \frac{m+2}{2^m}$ for all $m$ with $b_m\in C$. Notice that $\frac{n+2}{2^{n+1}}$ is not in any $b_m$ with $b_m\in C$. We conclude that $C$ is not a cover, so no finite subcover exists.

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    But is B an open cover of A? I figured it was but I do not know how to prove it – 2017-02-21
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    I added a proof that $B$ is a cover of $A$. – 2017-02-21
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    thank you very much for the help – 2017-02-21
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    Sure no problem. – 2017-02-21
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    Though I am curious about the second proof where n+2/2^n became n+2/2^n+1? – 2017-02-21
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    because the number on the right is smaller than $\frac{n+2}{2^{n}}$, so it is not contained in any of the intervals. – 2017-02-21
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Just theory not an technical answer.

We say that if a set of real numbers $A$ is compact then every open cover of $A$ has a finite sub-cover. This is one of several ways of characterizing compact sets in the real line. But, in the real line, we have that (True on real sets)

A set $A$ is compact iff $A$ is closed and bounded

A set $A$ is compact iff for every open cover of $A$ exists one finite sub-cover of $A$.

For this $A$ we have that $A$ is not closed, because we can construct a sequence that lies in $A$ such that converges to $0$, this implies that $A$ is not compact then not every open cover of $A$ will have a sub-cover.