# Thread: prove that u is open

1. ## prove that u is open

Prove that $U \subset M$ is open if and only if whenever $(Xn)^\infty_{n=1}$ is a sequence of elements of M converging to some $x \in U$, then there exists $N \in N$, such that $Xn \in U$for every $n>=N$.

this is one of the question in my midterm test that i couldnt do, the prof is not giving out solution, need some help to prove this

2. ## Re: prove that u is open

Originally Posted by wopashui
Prove that $U \subset M$ is open if and only if whenever $(Xn)^\infty_{n=1}$ is a sequence of elements of M converging to some $x \in U$, then there exists $N \in N$, such that $Xn \in U$for every $n>=N$.

this is one of the question in my midterm test that i couldnt do, the prof is not giving out solution, need some help to prove this
So, we're in a metric space, right? Assume first that $U$ is open but that this does not hold. Then, there exists $x\in U$ and a sequence $x_n\to x$ failing to eventually in $U$. Clearly then any neighborhood $x$ is going to contain all but finitely many of the $x_n$ and so, by assumption, an element of $M-U$. Thus, there does not exist a neighborhood of $x$ contained in $U$, which contradicts openess. Conversely, if $U$ is open and $x_n\to x\in U$ you know that there exists a neighborhood $V$ of $x$ with $V\subseteq U$. But, since $x_n\to x$ we have by definition that all but finitely many $x_n\in V\subseteq U$.