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The problem goes like: Suppose that we are given a point $x$ and a sequence $x_n$ in a metric space $M$, and suppose that $f(x_n) \rightarrow f(x)$ for every continuous real-valued function $f$ on $M$. Prove that $x_n \rightarrow x$ in $M$.

I was thinking that since the $f(x_n) \rightarrow f(x)$ for EVERY $f$, then we can find one with a continuous inverse (could it just be $f(x) = x$?). Then since both $f$ and $f^{-1}$ are continuous and also $f$ is bijective, it is a homeomorphism. Therefore, the preimage converges based on the fact that $f$ being a homeomorphism. Is this correct?

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  • We can't generally find a continuous $f$ with continuous inverse. If e.g. $M = \mathbb{Q}$, every continuous function $\mathbb{R}\to M$ is constant. – Daniel Fischer Oct 14 '15 at 15:49
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    The argument you proposed is not correct because $f(x)=x$ does not define a real-valued function and, more generally, there might not be any bijection between $M$ and the real line. I suggest defining $f$ by $f(z)=d(x,z)$, where $d$ is the given metric of $M$. – Andreas Blass Oct 14 '15 at 15:50

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$M$ is a metric space. Therefore it has a distance function, $d$, which is continuous if we fix one variable. Hence $$ d_x(y) := d(x,y) $$ is a continuous function in $y$. But then $d_x(x)=0$, and $x_n \to x$ iff $d_x(x_n) \to 0$.

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