# Cover of (0,1) with no finite subcover & Open sets of compact function spaces

I just got back from my exam and these questions' solutions eluded me, it would be great to use the rest of my evening figuring these out...

Q1: Find an open covering of the set $(0,1) \subset \mathbb{R}$, say $G =\{U_\alpha\}_{\alpha \in A}$, (where $A$ is some indexing set) such that $G$ has no finite subcover.

Q2: Let $f: [0,1] \to [0,\infty)$ be a continuous function. Let there be some $c\in [0,1]$ such that $f(c)$ is non-zero. Prove that there exists an $\epsilon \gt 0$ such that the set:

$X_1=\{\ x\in[0,1]\ | \ f(x)\gt\epsilon\ \}$

is non-empty, and open.

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Q1: think about the sets $(1/n,1)$ for $n\ge1$.

Q2: does it help if you take $\epsilon = f(c)/2$? Do you know something about preimages of open sets under continuous functions?

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For the first one, $\left\{\left(\frac1n,1\right):n\in\Bbb Z^+\right\}$ will do nicely.

For the second, let $\epsilon$ be any positive real number less than $f(c)$, and let $U=f^{-1}\big[(\epsilon,\to)\big]$. Since $(\epsilon,\to)$ is an open set in $\Bbb R$ and $f$ is continuous, $U$ is open in $[0,1]$, and the choice of $\epsilon$ ensures that $c\in U\ne\varnothing$.

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By $(\epsilon,\to)$ you mean $(\epsilon,\infty)$? – Pedro Tamaroff Dec 13 '12 at 0:31
@Peter: Yes; the arrow notation is also quite standard, and I prefer it. – Brian M. Scott Dec 13 '12 at 0:49
Yes, I think I'll start using it, too. Can I find you in chat? I'd like to tell you something. – Pedro Tamaroff Dec 13 '12 at 1:15
@Peter, Brian: I never saw this notation before Brian used it in the comments somewhere when we discussed linear orders. Of course that by that time I could infer the meaning of the symbol, but I don't know how standard it is. – Asaf Karagila Dec 13 '12 at 13:28

Because the obvious option was given (twice) for the first answer, let me give a cool alternative.

For $n>0$ let $a_n=\frac1n$. Now consider the intervals $(a_{n+1},a_n)$. Their union covers the set $(0,1)\setminus\{a_n\mid n\in\mathbb N^+\}$. Let $I_n$ be an interval which covers $a_n$ and is small enough (for what? read on to find out!).

Clearly $\{I_n\}_{n\in\mathbb N^+}\cup\{(a_{n+1},a_n)\}_{n\in\mathbb N^+}$ is an open cover of $(0,1)$. Argue that it is impossible to have a finite subcover.

If such finite subcover would exist then it would only contain a finite number of intervals of the form $(a_{n+1},a_n)$. This means that for some large enough $N$ we have that $(0,a_N)$ is contained completely in a finite number of $I_n$'s. Argue that $(0,a_N)$ cannot be contained in $\bigcup_{k=N}^\infty I_k$, because they are so small (i.e. their sum will never aggregate to $\frac1N$) and derive a contradiction.

Yes, it's much more to work with, but it's jolly fun and helps to understand the idea behind both measure zero sets and compactness.

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The questions were asked almost 4 years ago and don't know if there is any interest - had to say something about Q2 since I don't see anything redeemable about it. If we take the statement of the question at face value, the conclusion is wrong if the continuous function is a constant, $f(x)=k, k\in\mathbb{R}$. Choose $\varepsilon < k$, then $X_1=[0,1]$; choose $\varepsilon\geqslant k$, then $X_1=\varnothing$. Suppose $[0,\infty)$ is a co-domain spec; one would hope so since a continuous mapping with compact support can not generate an unbounded range; its range is closed and bounded. So say, $f:[0,1]\to[a,b]$ is continuous and onto with $b>a\geqslant0$. If we now suppose $f$ is one-to-one, the conclusion fails once again. The possible sets are $X_1=\varnothing$, $X_1=[0,1]$, $X_1=[0,p)$ or $X_1=(p,1]$ where $p\in(0,1)$ and $f(p)=\varepsilon$. The conclusion is true for some continuous functions that are not one-to-one. For example, $f(x)=1-x^2$ on $[0,1]$ fits the bill; choose $\varepsilon\in(0,1)$ and $X_1=(-\sqrt{1-\varepsilon},\sqrt{1-\varepsilon})$. The statement, in general, is true for those continuous functions that are not one-to-one, which attain exactly one maximum on $(0,1)$ or if they attain more than one maxima on $(0,1)$, they attain a specific value in $(a,b)$ an even number of times.

Your error is that you are not using the subspace topology on $[0,1]$. – Zev Chonoles Apr 2 at 20:15