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I have recently encountered this problem in my topology course. If $(X,\mathcal{T})$ is a topological space, show that if $X=A\cup B$ then $X=\bar{A}\cup B^{\circ}$, where $\bar{A}$ denotes closure and $B^\circ$ denotes interior. At the outset I thought this would be quite simple to prove, as intuitively it seems quite obvious, however that has not proved the case.

My thought was to just show containment in both directions. Showing that $\bar{A}\cup B^{\circ} \subseteq X$ is rather simple. The other way is providing headaches. I initially thought that contradiction would be the best attack so I assumed for some $x\in X$ that $x \notin \bar{A}\cup B^{\circ}$. Hence using the fact that in this case $A^c=B\backslash A$ and with the elementary properties of set operations I arrive at $x\in \bar{B}\backslash \bar{A} \cap \overline{A\backslash B}$. The inequality for the intersection of closures is the wrong way for me to conclude that $x\in \phi$ and so get my contradiction. I have tried tinkering for hours with different combinations of set differences and closures\interiors to no avail. I'm starting to think that contradiction isn't the best idea.

I have just started my topology course so I don't know more than the definition of a topology plus bases,closures, interiors, boundaries all defined in an abstract topological sense. This question lies before the notions of a metric and limit points have been introduced so I'm fairly certain that this question can be answered using simple techniques only. My other idea is to show that $\partial A= \partial B$ but I'm not sure where to start with this idea at the moment as my brain has retired hurt. If anyone can tell me which of my ideas to pursue and a push in the right direction I'd be very appreciative.

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  • $\begingroup$ Show the closure of the complement is the complement of the interior. $\endgroup$ – Jacob Wakem Aug 9 '16 at 0:27
  • $\begingroup$ @Alephnull I have already shown that and used it in my attempts to no avail unfortunately, $\endgroup$ – K.Power Aug 9 '16 at 12:23
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A direct proof, only based on the definitions:

All we have to prove is that an element $x$ of X which is not in $\mathring B$ is in $\,\overline{\!A}$.

If $x\notin\mathring B$, no neighbourhood of $x$ is contained in $B$, hence each neighbourhood meets $X\setminus B\subset A$, so each neighbourhood of $x$ meets $A$. This is the definition of $x\in\,\overline{\!A}$.

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If $A \cup B=X$, we know that $A^c \subset B$ (why?). Hence $(A^c)^o \subset B^o$. Thus it suffices to show $X=\bar{A} \cup (A^c)^o$, a much easier feat. To show $X \subset \bar{A} \cup (A^c)^o$, take $x \in X$, and suppose $x \notin \bar{A}$ (hence $x \notin A$, so $x \in A^c$), moreover there is an open set $U$ containing x such that $U \cap A= \emptyset$. Then $x \in A^c$, and the existence of this set $U \subset A^c$ containing x precisely says $x \in (A^c)^o$, so we're done.

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Lemma : If $(X,\mathcal T)$ is a topological space and $A\subseteq X$, then $\forall x\in X$ : $$x\in\overline{A}\iff \text{ every open subset of }X\text{ containing }x\text{ intersects }A.$$ Similarly $$x\in A^{\circ}\iff \text{ there is an open subset of }X\text{ containing }x\text{ which is included in }A.$$ (I let you prove it as an exercise.)

Now, let us prove that $X\subseteq \overline{A}\cup B^{\circ}$.

Let $x\in X$ : if $x\in B^{\circ}$, it's OK. Otherwise, $x\notin B^{\circ}$ : then (by the Lemma) there isn't any open subset of $X$ included in $B$ and containing $x$. Thus every open subset of $X$ containing $x$ intersects $A$ and (again by the Lemma) $x\in \overline{A}$.

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