I am asked to prove that given a topological space $T$, a set $G \subseteq T$ is open iff every point $x \in G$ has a neighborhood contained in $G$.

Now, this confused me because I am supposed to prove this just from the basic axioms, and I had previously seen a definition of an open set as this; I suppose I am confusing the topological axioms with the open sets we are used to in metric spaces.

Thanks in advance!


If the set $G$ is open then it is a neighborhood of each its point $x$. Conversely, if each point $x$ of $G$ has a neighborhood $U_x$ contained in $G$ then is has an open neighborhood $V_x$ contained in $U_x$, so the set $G'=\bigcup_{x\in G} V_x$ is open. Since each element $x$ of $G$ belongs to $V_x$, and $V_x$ is contained in $G'$, $G$ is contained in $G'$. On the other hand, each $V_x$ is contained in $G$, so in $G$ is contained also their union $G'$. So $G'=G$ and thus the set $G$ is open.

  • $\begingroup$ Why do you have to introduce V? Can't you just use U for the same thing? $\endgroup$ – Alis Sep 26 '17 at 16:33
  • $\begingroup$ @Alis In some definitions of a neighborhood $U$ of a point $x$ it may be not necessarily open, but still it should contain an open neighborhood $V$ of the point $x$. $\endgroup$ – Alex Ravsky Sep 26 '17 at 16:35
  • 1
    $\begingroup$ Got it, thanks! $\endgroup$ – Alis Sep 26 '17 at 16:36

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