Let $f : [0,1] \rightarrow \mathbb{R}$ be continuous. Suppose that $(x_n)_{x \in \mathbb{N}}$ is a sequence in $[0,1]$, with $f(x_n) \rightarrow 0$. Show that there exists $\psi \in [0,1]$ with $f(\psi) = 0$

Do we show that $\psi$ can be found within the sequence. And if the sequence tends to zero, then we know that all sub-sequences tend to the same value?

  • $\begingroup$ Hint: $y_n=f(x_n)$ is a convergent sequence in the compact set $f([0,1])$. $\endgroup$
    – dxiv
    Oct 21, 2016 at 19:53
  • $\begingroup$ it is not true that $\psi$ can always be found in the sequence. Take e.g. $f(x)=x$ and $x_n=1/n$. $\endgroup$
    – TonyK
    Oct 21, 2016 at 19:57

1 Answer 1


Extract from $x_{n}$ a convergent subsequence $x_{n_{k}}$ (which necessarily exists because $[0, 1]$ is compact). Denote the limit of $x_{n_{k}}$ by the symbol $\psi$. This limit lies in $[0, 1]$ because the interval is closed.

This proof is not quite complete, but almost... :)


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