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Thread: prove that u is open

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    prove that u is open

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

    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
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    Re: prove that u is open

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

    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 $\displaystyle U$ is open but that this does not hold. Then, there exists $\displaystyle x\in U$ and a sequence $\displaystyle x_n\to x$ failing to eventually in $\displaystyle U$. Clearly then any neighborhood $\displaystyle x$ is going to contain all but finitely many of the $\displaystyle x_n$ and so, by assumption, an element of $\displaystyle M-U$. Thus, there does not exist a neighborhood of $\displaystyle x$ contained in $\displaystyle U$, which contradicts openess. Conversely, if $\displaystyle U$ is open and $\displaystyle x_n\to x\in U$ you know that there exists a neighborhood $\displaystyle V$ of $\displaystyle x$ with $\displaystyle V\subseteq U$. But, since $\displaystyle x_n\to x$ we have by definition that all but finitely many $\displaystyle x_n\in V\subseteq U$.
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