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I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.

The Prompt:

Let $K \subset R^1$ consist of $0$ and the numbers $\frac{1}{n}$, for $n = 1,2,3,\dots$. Prove that $K$ is compact directly from the definition (without using the Heine-Borel theorem).

My Proof:

Define a sequence $(k_n)$ where $k_0 = 0$ and $k_n = \frac{1}{n}$, so that $\cup(k_n) = K$. Suppose $\{G_\alpha\}$ is an open cover of $K$. Then $\cup_{n=0}^{\infty}B_r(k_n)\subset \cup\{G_\alpha\}$ for infinitely many open balls of radius $r$. But, for any $r$, $B_r(k_0) = B_r(0)$ will contain infinitely many elements of $(k_n)$ as shown:

$(k_n) = \frac{1}{n} < 0 + r = r \to \frac{1}{r} < n$

So, all values of $(k_n)$ where $n > \frac{1}{r}$ are contained in $B_r(0)$. Thus, $\cup_{n=0}^{\frac{1}{r}}B_r(k_n)$ covers $\cup(k_n)$ and $\cup_{n=0}^{\frac{1}{r}}B_r(k_n)\subset \cup\{G_\alpha\}$. Therefore, it is a finite subcover of $K$ and so $K$ is compact.

user10360304
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1 Answers1

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Elements that are lacking clarity in your proof:

  • You haven’t define what the $B_r(k_n)$ are.
  • Reading between the lines, we can suspect that those are open balls with radius equal to $r$ centered on $k_n$.
  • Which leads to following issues: (1) an open cover maybe by something else than balls, (2) the balls may have different radius and (3) may not be centered on the $k_n$.

For a proof, I would say that if an open cover $\mathcal U$ covers $K$, then it will exists an open $U \in \mathcal U$ such that $0 \in U$. Then prove that $U$ contains all but a finite number of elements of $K$. Conclude from there.

mathcounterexamples.net
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  • I totally agree on the clarity comment, however I don't quite get the issues mentioned in the third bullet point. The open cover certainly could be made by something other than balls, and the balls might have different radii or may not be centered on the $k_n$, however, wouldn't there always have to be an infinite set of balls that are a subset of the open cover? And then my proof created a finite subcover of that subset of the open cover (which would also be a finite subcover of the full open cover). – user10360304 Apr 24 '20 at 20:10
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    No, the balls will not necessarily be a subset of the open cover. However, for each open, you could find a ball that is included in that open of the open cover and that contains a point of $K$ that is itself in an open of the open cover. – mathcounterexamples.net Apr 24 '20 at 20:15
  • Sorry, I'm still not getting it. Even if I define $r$ as the minimum distance from any $k_n$ to the edge of some part of the open cover, an infinite set of balls of radius $r$ wouldn't be a subset of the cover? Could you maybe share an example of when a set of open balls would not be a subset of the open cover? – user10360304 Apr 24 '20 at 20:26
  • $r$ defined as you suggest may be equal to zero. – mathcounterexamples.net Apr 24 '20 at 20:31
  • Sorry, yeah that's right. What if I define $r$ as the minimum distance from any $k_n$ to the edge of the union of the open cover? Then, r couldn't be zero because then that would imply $k_n$ was not in the open cover, which would make it not an open cover of $K$. – user10360304 Apr 24 '20 at 20:55