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I am just getting into topology, and I have a doubt regarding open sets.

Let $(X, \mathcal{T})$ be a topological space. Given an open set of $X$, $A$, and subset of $X$, $B$ such that

$$A\cap B \in \mathcal{T}$$ $$A\cup B \in \mathcal{T}$$

Can I conclude that $B$ is also an open set? That is, if I have an arbitrary set of $X$ whose intersection and union with an open set are themselves open sets, does this imply the arbitrary set is also open?

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  • $\begingroup$ How about this example where $A\cap B\ne\emptyset$ and $A\cup B\ne X$: $X=\{1,2,3,4\}$, $\mathcal T=\{\emptyset,\{1\},\{2\},\{1,2\},\{1,2,3\},\{1,2,3,4\}\}$, $A=\{1,2\}$, $B=\{2,3\}$ $\endgroup$ – J. W. Tanner Jun 7 '20 at 11:10
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No. It does not imply that the set is open. For instance:

Let $X= \left\lbrace a, b, c\right\rbrace$ and consider the topological space $(X, \tau)$ where $\tau=\left\lbrace\varnothing, X, \left\lbrace a \right\rbrace\right\rbrace$.

Let $A=\left\lbrace a \right\rbrace$, let $B=\left\lbrace b, c \right\rbrace$.

Then,

$A \cup B= X \ \in \ \tau $,

$A \cap B= \varnothing \ \in \ \tau$.

However $B \ \notin \ \tau$.

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    $\begingroup$ I would recommend denoting the empty set as $\emptyset$ rather than $\phi$. $\endgroup$ – Zest Jun 4 '20 at 16:24
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    $\begingroup$ You can use either \emptyset $\emptyset$ or \varnothing $\varnothing$. $\endgroup$ – Paul Sinclair Jun 5 '20 at 3:02
  • $\begingroup$ True that! Thanks for the recommendations. $\endgroup$ – matumath Jun 5 '20 at 14:53
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No. Consider the usual topology on $X=\mathbb R$ with $A=(-\infty,0)$ and $B=[0,\infty)$.

$A\cap B=\emptyset\in\mathcal T$, $ A\cup B=X\in\mathcal T$, $A\in\mathcal T$, and $B\not\in\mathcal T$.

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