This might be a silly question; this is where I'm stuck as to whether a metric-bounded set in a connected metric space is uniformity-bounded in the sense of Bourbaki. Let $(X,d)$ be a bounded connected metric space, $x_0\in X$ and $\epsilon>0$. Let $(A_n)_{n=0}^\infty$ be an increasing sequence of subsets of $X$ defined by $$A_0=\{x_0\}$$ $$A_n=\bigcup_{x\in A_{n-1}}B(x,\epsilon)\qquad(n>0).$$ The union $A=\bigcup_{n=0}^\infty A_n$ of the sets is obviously open and not empty. Since $B(x,\epsilon)\cap A_n\ne\varnothing$ implies $x\in A_{n+1}$, $A$ is also closed. Therefore we have $A=X$ since $X$ is connected. My question is: is there always an $n$ for which $A_n=X$?
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$\begingroup$ You can prove by induction that $A_n = B(x_0 , n \epsilon)$. Since $X$ is bounded, at some point that ball will contain all points of $X$. $\endgroup$– CrostulCommented Jul 5 at 22:37
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$\begingroup$ @Crostul I don't think your first statement is true. $\endgroup$– arkeetCommented Jul 5 at 22:44
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This is false. Consider $\mathbb{R}$ with the metric $d(x,y) = \min(|x-y|, 1)$, and $x_0 = 0$. This is clearly bounded, but if $\epsilon < 1$, we have for $n > 0$ that $A_n = \{x : |x| < n\epsilon\} \ne \mathbb{R}.$
(A sufficient condition would be for the space to be totally bounded.)
