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I don't understand the argument at 26:00-27:14. Where did he use the definition of the standard $\mathbb{R}^d$ topology? It seems that the empty set is in any family of sets by that proof which is clearly nonsense.

A youtube snapshot is helpful - compare the definition of $\mathcal U$ on the left side of the board to the right side where he proves $\emptyset \in \mathcal U$.

enter image description here

CopyPasteIt
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nonuser
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    I couldn't follow all of it because I find the definition of "the standard R[sup]d[/sup] topology". But it appears that what he proves (at time 26 minutes into the lecture) is that the empty set is a subset of any set, not that it is a member of a specific collection of sets. – user247327 Feb 02 '20 at 15:22
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    I would realy like to see some explanation before the down vote. What is wrong with the question? – nonuser Feb 02 '20 at 15:36
  • @user247327 This is not what he proves here. I'm not sure why you think that. He proves that the empty set is an element of the topology. – G. Chiusole Feb 02 '20 at 15:53
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    I didn't downvote, but I would expect someone of your reputation to do your best to make questions self-contained. If that video goes down, this question becomes far less useful to others. – Mark S. Feb 02 '20 at 16:01
  • The proof shows that the empty set belongs to every family $\mathscr{F}$ of sets that is defined by a condition $S \in \mathscr{F} \iff \bigl(\forall p \in S : \text{whatever}\bigr)$. – Daniel Fischer Feb 05 '20 at 21:30
  • I'm talking about the indicated part of the video that you linked. $\mathcal{O}{\text{st}}$ is defined by $U \in \mathcal{O}{\text{st}} \iff \bigl(\forall p \in U : \exists r \in \mathbb{R}^+ : B_r(p) \subseteq U\bigr)$. The proof shows — though that's not quite explicitly stated — that the empty set belongs to every family of sets defined by such a condition. (Because the part after "$U \in \mathcal{O}_{\text{st}} \iff \bigl(\forall p \in U :$" isn't used at all; $\forall p \in \varnothing : \ldots$ is vacuously true.) – Daniel Fischer Feb 05 '20 at 21:51
  • So if I define some family $F$ of sets with property $P$ then $\emptyset$ is in this family because $∀p∈\emptyset$:… is vacuously true @DanielFischer – nonuser Feb 05 '20 at 22:02
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    If $P(S)$ is of the form $\forall p \in S : \ldots$, then yes. If the property has a different form, then $\varnothing$ may or may not have that property. – Daniel Fischer Feb 05 '20 at 22:06
  • Yes, this make sense now. He actually proved that emptyset is open for much larger class of families no just family made in standard $R^d$ topology. He could mentioned that and there would be no confusion (for me of course). Also proofs with contradiction (which are equivalent to EFQ) are much easier to understand. Thanks@DanielFischer Can you please write this comments as an answer. – nonuser Feb 05 '20 at 22:28
  • This reminds me of the fact that the number of bijective mappings between the empty set and itself is equal to $1$. Can anyone say $0! = 1$. – CopyPasteIt Feb 06 '20 at 03:19
  • See also Base and empty set of a topology. – CopyPasteIt Feb 06 '20 at 03:29

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On the left hand side of the board, the standard topology on $\mathbb{R}^d$ is defined by: A subset $U \subseteq \mathbb{R}^d$ is in the standard topology, if and only if $\forall p \in U: \exists r > 0$ s.t. $B_r(p) \subseteq U$.

With this definition, the empty set is indeed in the standard topology. We just need to check that $\forall p \in \emptyset: \exists r > 0$ s.t. $B_r(p) \subseteq \emptyset$. However, this is trivially true, since there are no $p \in \emptyset$.

More precisely, the statement is true since we have

$$\forall p,\exists r > 0: p \in \emptyset \Rightarrow B_r(p) \subseteq \emptyset$$

However, this means that for every $p$ we have $\text{false} \Rightarrow \text{something}$, which by Ex falso quodlibet, is a true sentence.

Also, the proof does not show that the empty set is in any family of sets. This is not true. For example, consider the following collection of subsets of $\mathbb{N}$ defined via $S:= \{U \subseteq \mathbb{N} \vert 1 \in U\}$. It does not contain the empty set.

However, the proof does show that for any collection $A$ of sets defined by $U \in A$ if and only if $\forall p \in U: \text{something}$ the empty set is a member.

G. Chiusole
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