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Let $X$ be a topology space and $A$ be a subset of $X$. For each $x\in A$, there is an open set $U$ containing $x$ such that $U$ is a subset of $A$. Show that $A$ is open in $X$.


Let $\mathcal{B}$ be a basis for $X$. By definition, since $U$ is open there exist $B\in \mathcal{B}$ such that for $x\in U$, $x\in B \subset U$ and since U is a subset of $A$, $x\in B \subset A$. So, $A$ is open.

My only concern is that we start with an $x \in U$ and $U$ is a subset of $A$ but may not contain all the elements of $A$. So, not all elements in $A$ will lie in $B$. Is that true?

I use the definition: A subset $U$ of $X$ is said to be open in $X$ if for each $x∈U$, there is a basis element $B∈B$ such that $x∈B$ and $B⊆U$.

Is my solution correct?

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Yes. If you're definition of open set is "For every $x\in A$ there is a basis element $B \in \mathcal{B}$ such that $x\in B$ and $B\subset A$" – Prahlad Vaidyanathan Oct 4 '13 at 16:21
What definition of open set do you use? – Stefan Hamcke Oct 4 '13 at 16:27
@StefanH.: I have included the definition – James Bond Oct 4 '13 at 16:36
Usually you don't need a basis for that. – André Caldas Oct 4 '13 at 17:55

It seems to me a bit weird to use the notion of basis here. The characterisation of a basis is at least as hard as a direct proof of this.

I would prove this purely based on the definition of a topological space. There are various competing definitions, but the one most often used is: a topological space is a set $X$ equipped with a collection ${\cal U}$ of subsets of $X$ satisfying: (1) $\emptyset, X \in {\cal U}$; (2) ${\cal U}$ is closed under arbitrary unions; (3) ${\cal U}$ is closed under finite intersections. The sets in ${\cal U}$ are called the open sets of $X$.

Proposition. Let $X$ be a topological space and let $A$ a be subset of $X$ with the property that for every $x$ in $A$ there is an open subset $U$ of $X$ with $x \in U \subseteq A$. Then $A$ is open.

Proof. Let $I$ be the union of all open subsets of $X$ that are contained in $A$ (the interior of $A$, commonly denoted by $A^\circ$). By construction, $I \subseteq A$, and by (2) $I$ is open.

Take $x \in A$. By assumption, there is an open subset $U$ of $X$ with $x \in U \subseteq A$. By construction of $I$, $U \subseteq I$ and therefore $x \in I$. This shows that $A \subseteq I$ as well.

Therefore $A = I$ and hence $A$ is open.

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Your solution is correct, but I would formulate it differently:

Let $B$ be a basis for $X$. By definition, since $U$ is open there exist $B∈\mathcal B$ such that for x in U, $x∈B⊂U$ and since $U$ is a subset of $A$, $x∈B⊂A$. So, $A$ is open.

since the $B$ depends on the $x$.

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I would formulate it with $\:\subseteq\:$ instead of $\:\subset\;$. $\;\;\;$ – Ricky Demer Oct 4 '13 at 18:13

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