# Condition of connectedness

Is it true that two subsets of a metric space are disconnected if and only if there does not exist a Cauchy sequence with all points in one subset and its limit in the other?

It seems right for simple cases, though I can't come up with a rigorous proof.I'm having a hard time with the concept of connectedness.

Edit:
Is it true that a metric space is connected if and only if for each proper subset $A$, there exists another subset $B$, connected to $A$, such that $A \cap B = \emptyset$?

Thanks.

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You should have stated the definition of "disconnected pair of subsets" that you are using. If it means $\overline A\cap B=A\cap\overline B=\varnothing$, then yes, this is the equivalent to the property with the Cauchy sequences that you stated. – user31373 Jun 2 '12 at 18:58
I'm not sure what definition I'm using. Yours looks nice and the equivalence is obvious. I asked this because the question of whether some metric space is connected often boils down to the question whether some two subsets of it are connected and I didn't see a clear way to tell. – Karolis Juodelė Jun 2 '12 at 19:12

Let $(X,d)$ be our metric space.
I'll be using the definition of connectedness where a space is disconnected if there are open $\emptyset\neq A, B\subset X$ such that $A\cup B=X$ and $A\cap B=\emptyset$.
Since our sequence $\left(x_n\right)_{n\geq0}$ converges by assumption, we can find, for any open $U\subset X$ with $x\in U$, an $N$ such that $x_n\in U$ for all $n\geq N$ (this definition of convergence is equivalent with the $\varepsilon$-$N$ one).
Assume $(X,d)$ is disconnected, and all $x_n$ are in $B$, and $x\in A$. Since $A$ is open, surely we can find $N$ such that $x_n\in A$ for all $n\geq N$, which is nonsense.