A function $f: \mathbb R \to \mathbb R$ is called sequentially continuuous if

$x_n \to x$ implies $f(x_n) \to f(x)$.

Every continuous function is sequentially continuous. Let $f$ be a continuous function on $\mathbb R$.

If $y_n = f(x_n)$ is a convergent sequence does it follow that $x_n$ is a convergent sequence?

At first I thought that yes because: assume $f(x_n) \to f(x)$ but $x_n \not\to x$. Say, $x_n \to z$. Then since $f$ is sequentially continuous, $f(x_n) \to f(z)\neq f(y)$. Although $x_n \not\to x$ is not equivalent to $x_n$ converging to a different value it seems that the case where $x_n$ diverges can be treated similarly.

But then I came up with an almost counterexample: Let $x_n =n$ and $f(x) = {1\over x}$. Then $f(x_n) \to 0$ but $x_n \to \infty$. The problem is, this $f$ is not defined on $\mathbb R$ and also, there is no $x$ with $f(x_n) \to f(x)$.

Is it possible that $f(x_n) \to f(x)$ implies $x_n \to x$?

  • $\begingroup$ If this condition holds, then $f$ is 1-1, and $f^{-1}$ (defined on the range of $f$) would be continuous. $\endgroup$ – Prahlad Vaidyanathan Mar 8 '15 at 2:56

Probably not. Example: $f(x) = x^2$ is continuous on $\mathbb{R}$, and take $x_n = (-1)^n$ does not converge, but $y_n=f(x_n) = 1, \forall n$, hence converges to $1$. Another example would be to take $f(x) = |x|$, and the same $x_n = (-1)^n$ would work just fine.


Take any constant function. Then $y_n$ is trivially convergent for any $x_n$.

  • $\begingroup$ Just to clarify, that is a counterexample, so the answer to OP's question is no. Right? $\endgroup$ – evaristegd Jun 16 '19 at 23:30
  • 1
    $\begingroup$ @evaristegd: Correct. $\endgroup$ – copper.hat Jun 17 '19 at 0:44

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