# An equivalent characterisation of open subset of a topological space

I'm having trouble understanding topologies.

We say that $U \subseteq X$ is open if $U \in \tau$. If $(X, \tau)$ is a topological space and $U \subseteq X$, why are these properties the same?

1. $U$ is open
2. For each $x \in U$, there is an open $U_x$ with $x \in U$ and $U_x \subseteq U$.
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You need to also understand the axioms of topology. $\tau$ is not just some arbitrary collection of subsets, it has a particular structure. – JHance Jul 20 '14 at 19:30

If $U$ is open, $2$ trivially holds with $U=U_x$ throughout. If $2$ holds, $U=\bigcup\limits_{x\in U} U_x$ is a union of open sets, so it is open.

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Why does U = $U_{x}$ trivially hold if U is open? – user164179 Jul 20 '14 at 17:48
I didn't say that. I said that if $U$ is open, you can simply take $U=U_x$ for any $x\in U$, since $U$ is open and $U\subseteq U$ to prove $1\implies 2$. – Pedro Tamaroff Jul 20 '14 at 17:50

From 1 to 2: we pick $U_x = U$ for every $x \in U$.

from 2 to 1: assuming we have such $U_x \in \tau$ for every $x$, $U = \bigcup \{U_x: x \in U\}$.

(because every $x \in U$ is in its "own" $U_x$, so in the union, which shows one inclusion and because every $U_x \subset U$, so their union as well, which shows the other.)

Then $U$ is a union of members of $\tau$, hence in $\tau$ itself.

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