I am going through Rudin's Functional Analysis 1st chapter. While doing a proof in Topological vector space he used $V+\emptyset=\emptyset$, where $V$ is a open set and $\emptyset$ is empty set in the topological vector space. But I think it should be $V$. Have I misunderstood the addition?

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    $\begingroup$ Little note on MathJax: for the empty set, you can use "\emptyset" - of course within Dollar symbols: $\emptyset$ $\endgroup$ – haemi Aug 25 '17 at 12:15
  • $\begingroup$ No, you've not really misunderstood anything serious. This is an "edge case" that can only be parsed in a formal sort of way, as clarified by @user275313's correct answer. It's not a serious issue... $\endgroup$ – paul garrett Aug 25 '17 at 21:22

The claim in the book is correct. If $A$ and $B$ are two subsets of a vector space, then we define $A+B$ as $$A + B = \{a+b \mid a \in A, b \in B\}$$

In your case, $B$ is the empty set so there are no $b \in B$ for you to be forming $a+b$ terms, so $A+B$ is the empty set.

You might be thinking of the case where $B$ is $\{0\}$, i.e. the set containing only the zero vector. In that case

$$A + B = \{a+b \mid a \in A, b \in \{0\}\} = \{a+0 \mid a \in A\} = A$$


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