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Problem. Let $X$ be a non-empty set and $\beta$ be a basis set (in the linked definition of basis for a topology replace the term by basis set). Then the set $\tau_\beta:=\left\{\displaystyle\bigcup_{B\in \beta_0} B: \beta_0\subseteq \beta\right\}$ forms a topology on $X$.

I have already shown that $\tau_{\beta}$ is a topology assuming that $\emptyset\in\tau_{\beta}$ but I can't show that $\emptyset\in \tau_\beta$. I think that it holds vacuously but I am not sure how to prove it. Can anyone help?

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$\varnothing$ is a subset of $\beta$ and is therefore a possible $\beta_0$, so one of the elements of $\tau_\beta$ will be $\displaystyle\bigcup_{B\in\varnothing} B$ which is empty.

hmakholm left over Monica
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  • So, do we define $\displaystyle\bigcup_{B\in\emptyset} B=\emptyset$ or it is a vacuous truth? –  Jul 09 '16 at 14:40
  • What are the elements of $\bigcup_{B\in\emptyset} B$? There are none, so it is the empty set. You can call this vacuous truth or a definition, or something proved, depending on the system you are working in, but it is the only meaning that makes sense. @user170039 – Thomas Andrews Jul 09 '16 at 14:43
  • @user170039: The definition of $\bigcup\limits_{B\in A} B$ is, as always, the set of all things that are elements of some element of $A$. In the case where $A=\varnothing$, then this works out to the set of no things, because there is nothing that is an element of an element of $A$ when $A$ has no elements. – hmakholm left over Monica Jul 09 '16 at 14:46
  • This is the intuition behind defining the union to be $\emptyset$. But my question is do we actually define the union to be $\emptyset$ or it can be proved? For if it's just a matter of definition, we could as well define it to be something else, say the whole set $X$. –  Jul 09 '16 at 14:56
  • @user170039: You can define anything to mean anything you want, but if you want to be able to communicate with people, then you had better use the standard definitions, or something equivalent to them. The standard definition of $\bigcup_{B\in A} B$ is the set whose members is everything that is an element of an element of $A$ and nothing else. If you change that definition such that in means something else than this when $A$ happens to be the empty set, then that's your prerogative, but it's on yourself when subsequently nobody understands what you're saying. – hmakholm left over Monica Jul 09 '16 at 15:00
  • I think that I was not clear enough in expressing my question, so let me try once again. My question is, do we actually define the union to be $∅$ or it can be proved (e.g., shown to be vacuously true)? –  Jul 09 '16 at 15:03
  • @user170039: Let me try once again. We actually define $\bigcup_{B\in A}B$ to mean the set of everything that is an element of an element of $A$ and nothing else. That is the standard definition, and the full and complete definoition. This is not a matter of proof, no matter whether $A$ is empty or not -- it is simply what the notation means. If you want $\bigcup_{B\in\varnothing}B$ to mean something different from what this standard definition says, then you're using an extremely non-standard definition, and you deserve no pity when you find you cannot communicate with people. – hmakholm left over Monica Jul 09 '16 at 15:07
  • Do you need me to substitute $A=\varnothing$ into the definition for you? Very well, by the above standard definition, $\bigcup_{B\in\varnothing}$ means the set of everything that is an element of an element of $\varnothing$, and nothing else. This is the empty set because every single thing in the world is not an element of any element of $\varnothing$. So every single thing in the world is a non-element of $\bigcup_{B\in\varnothing}B$. Nothing is an element of it. It is a set with no elements. It is an empty set. THIS IS AN EX-PARROT. – hmakholm left over Monica Jul 09 '16 at 15:13
  • Thank you very much. Actually I was reading from Munkres's book. There it is written that, "There is no problem in these definitions [the definitions of arbitrary union and arbitrary intersection] But it is a bit tricky to decide what (if anything) these definitions mean if we allow $\mathcal{B}$ to be an empty collection. Applying the definitions literally, we see that no element $x$ satisfies the defining property for the union of elements of $\mathcal{B}$. So it is reasonable to say, $$\displaystyle\bigcup_{B\in\mathcal{B}}B=\emptyset$$ if $\mathcal{B}$ is empty." –  Jul 09 '16 at 15:17
  • The bold part of the quote made me think that defining $\displaystyle\bigcup_{B\in\emptyset}B=\emptyset$ is more or less arbitrary. –  Jul 09 '16 at 15:18
  • @user170039: It would definitely be arbitrary to define $\cup_{B\in\varnothing}B$ as a special case. It should be defined by the standard definition which I have given several times now and am tired of retyping. Claiming it to be defined by a special case that leads to the same result as the general definition it purports to be an exception to, is worse than arbitrary -- it is arbitrary and extremely misleading. – hmakholm left over Monica Jul 09 '16 at 15:24
  • Yeah. That's what I thought too. Munkres's writing is very much confusing at places. –  Jul 09 '16 at 15:26