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$\forall t\in I=[0,1]$, choose an open path-connected neighborhood $V_t$, then $\{V_t\}$ is an open cover of $I$. Does $\{V_t\}$ have a finite subcover $V_{t_1},V_{t_2},\cdots,V_{t_n}$ such that $0=t_1<t_2<\cdots<t_{n-1}<t_n=1$ and $V_{t_i}\cap V_{t_{i+1}}\neq \varnothing$ for $i=1,2,\cdots,n-1$?

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  • $\begingroup$ What do you think? $\endgroup$ – mrp Mar 17 '15 at 7:50
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Yes. You can prove it by modifying a standard proof that $I$ is compact.

HINT: Say that $t\in I$ is good if there is a finite sequence $0=t_0<t_1<\ldots<t_n=t$ such that $[0,t]\subseteq\bigcup_{k=1}^nV_{t_k}$, and $V_{t_k}\cap V_{t_{k+1}}\ne\varnothing$ for $k\in\{0,\ldots,n-1\}$. Let Let $A=\{t\in I:t\text{ is good}\}$.

  • Show that $A$ is bounded above and non-empty. Conclude that $\sup A$ exists.
  • Show that $\sup A\in A$.
  • Show that $\sup A=1$.
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