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I know it's true in case of open interval and open covering.

Also, in this case what will be Lebesgue number ?

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$I = [0, 1]$ is closed as a subset of itself, making $\{ I \}$ a closed covering of $I$.

$B_{\delta}(x) \cap I \subset I$ for each $x \in I$ and each $\delta \in \mathbb{R}_{> 0}$, so that every such $\delta$ is a Lebesgue number.

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  • $\begingroup$ Yes but won't $\delta$ tends to zero ? As we take x near to 0 or 1, we have to take $\delta$ smaller correspondingly. $\endgroup$ – MeetR Apr 18 '17 at 2:52
  • $\begingroup$ The open balls in $I$ as a metric space are all of the form $B_{\delta}(x_0) \cap I = \{ x \in I : |x - x_0| < \delta \}$ for $x_0 \in I$. $\endgroup$ – Dean Young Apr 18 '17 at 2:57

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