If $m\in\mathbb{N}$, $M=\{0,1\}^{\mathbb{N}}$ and $A \subseteq M$ is an open set of sequence where the number 1 appears at least $m$ times.
Show that $A$ is an open subset of $M$.
I wanted to show that if there exists a circle with center in $x$ that is $K(x,r) \subseteq A$ then $A$ will be an open subset. I'm not sure if that is the case though or how to show it.
EDIT: I know that
$$\mu (x,y) = \min\{n\in\mathbb{N} \ | \ x_n \not= y_n \}$$
and
$$d(x,y) = \frac{1}{\mu(x,y)}$$
and the metric space is given by $(M,d)$
First, let's get your terminology straight. You need to be more precise if you want to write proofs correctly. You're introducing the letters $x$, $r$ and $K$ without defining what they mean, and while you can usually infer some information from a name and context, it's better to be slightly more wordy. (Also, when asking questions like these: specify the topology! But you fixed that in the edit)
I'm assuming you want to use the criterion that a subset of a metric space $A \subseteq M$ is open if and only if for each point $x \in A$, we have that the open ball centred at $x$ with radius $\varepsilon$ $$ B(x, \varepsilon) = \{ y \in M : d(x, y) < \varepsilon \} $$ is fully contained in $A$.
So, first take any point $x$ in $M$. What does $B(x, \varepsilon)$ look like for different $\varepsilon$?
Once you've figured that out, take any point $x \in A$. Can you choose an $\varepsilon$ such that $B(x, \varepsilon)$ will be fully contained in $A$? Use the special property that you know about $A$.