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Say we have the following open cover of $(0,1)\subset\mathbb{R}$:

$$O=\bigcup_{n=2}^\infty \left(\frac{1}{n},1\right)$$

So this an open cover of $(0,1)$, but could I not turn it into an open cover of $[0,1]$ by defining the following set $S$:

$$S=O\cup (-\epsilon,\epsilon)\cup(1-\epsilon,1+\epsilon)$$

For some $\epsilon>0$.

Is this an open cover of $[0,1]$? If not why? If so how can I find a finite subcover of $S$ that still covers $[0,1]$?

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Yes, $S$ is an open cover of $[0,1]$. Choose $n\in\Bbb Z^+$ large enough so that $\frac1n<\epsilon$; then

$$\{(-\epsilon,\epsilon),(1-\epsilon,1+\epsilon)\}\cup\left\{\left(\frac1k,1\right):1\le k\le n\right\}$$

is a finite subset of $S$ covering $[0,1]$.

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