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In Munkre's topology, there are two instances where we would have to represent the empty set as a union of sets.

First, we have a lemma that says if $X$ is a set; and $\mathcal{B}$ is a basis for a topology $\tau$ on $X$, then $\tau$ is equal to the collection of all unions of elements of $\mathcal{B}$.

Later on, we define a subbasis $S$ of a topology on a set $X$ to be a collection of subsets of $X$ whose union equals $X$, and the topology generated is the collection of all unions of finite intersections of elements of $S$.

So there are a few questions I wanted to ask about these just to make sure that I'm understanding them correctly. First in the case of the subbasis, when Munkres says finite intersections, he's allowing for a set to be intersected with itself right? There's nothing that says the sets have to be different? This is important because we will need the elements of S to be in the collection of finite intersections so that we can get the full set X.

Secondly, and more importantly, when dealing with the empty set, how do we say that this set is a union of anything? If I have a collection of sets D and I take the empty collection $\mathcal{A}$, then can I say the union of the empty collection is the empty set? I think this is ok to do, and Munkres mentions this earlier in the book, so is he allowing this to occur when he says all unions of elements from a collection of sets? I suppose its not that hard to understand but I guess I'm not used to thinking this way.

Lastly, in the proof of the lemma I mentioned first, Munkre's proves that every open set is a union of basis elements in the following way. Let $U \in \tau$, choose for each $x \in U$ an element $B_x$ of the basis such that $x \in B_x \in U$. Then $U = \bigcup_{x \in U} B_x$, so $U$ equals a union of elements of the basis.

I'm having trouble interpreting this statement if we start with $U=\emptyset \in \tau$. Then there is no $x \in U$, so how do we interpret the rest of the statement?

Thank you

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The nullset is the union of a set of 0 things, and since there are no sets for any element $x$ to be in, statements about such an $x$ are vacuously true.

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  • $\begingroup$ In particular, in the last statement you have chosen no $B_x$es, so you take a union of no sets to get $\emptyset$. $\endgroup$ – Kevin Carlson Mar 7 '16 at 21:14

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