Prove that $F$ is closed

Question

We are given that if a sequence $(x_n)\subset F$ converges to some point $x\in M$, then $x\in F$. We must prove that $F$ is closed.

My attempt:

I strongly believe that the trick here is to come up with a special sequence in $F$ that converges to a point $x\in M$. Then we can use the hypothesis to say that $x\in F$, which will presumably help us prove the desired result, i.e., $M\setminus F$ is open.

Could someone give me a hint?

Answer 1

Suppose $F$ is not closed. Then there is a point $x \in M$ such that $x \notin F$ but every open ball (I'm assuming a metric space situation, as mentioned in my comment) around $x$ intersects $F$. Pick $x_n \in B(x, \frac{1}{n}) \cap F$ and show that $x_n \to x$, which contradicts the assumption on $F$.

Answer 2

Assume $M\setminus F$ is not open. Then there is a point $x$ in $M\setminus F$ such that every neighbourhood of $x$ has nonempty intersection with $F$.

In particular this holds for the balls with radius $n$ centered at $x$. For every natural number $n$ choose $x_n$ out of the intersection between the ball with radius $n$ centered at $x$ and $F$.

Clearly $x_n$ converges to $x$. The condition on $F$ yields that $x$ is an element of $F$. Contradiction.

Answer 3

Let $y\in M$. If, for every $n>0$, $B(y,1/n)\cap F\ne\emptyset$, you can choose $x_n\in B(y,1/n)\cap F$. The sequence $(x_n)$ converges to $y$ (prove it). Hence $y\in F$.

Therefore, if $y\in M\setminus F$, there exists $n>0$ with $B(y,1/n)\cap F=\emptyset$, so $B(y,1/n)\subseteq M\setminus F$. Hence $M\setminus F$ is open.

Answer 4

Twisted idea: define $$f(x) = d(x,F) = \inf\{d(x,y)\,:\,y\in F\}.$$ You can check easily that $f$ is continuous. Now, you can use your hypothesis to prove that $$x_0\in F\iff f(x_0) = 0$$ (suppose $f(x_0) = 0$ and take a sequence of $x_n\in F$ s.t. $d(x_0,x_n) < 1/n$).

Finally, $F$ is the inverse image of a closed set: $F = f^{-1}(\{0\})$.