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"If $U$ is an open set in normed linear space $X$, then for any $x\in X$, $x+U$ is open". My question is: What can we say if $X$ is not a normed linear space, will it hold? I tried finding counterexamples but couldn't find it. Any hints?

Lucas
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    What is the definition of $x + U$ in a non-linear space? – Didier Dec 09 '20 at 15:02
  • $x+U ={x + u : u\in U}$ – Lucas Dec 09 '20 at 15:03
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    What is $x+u$ in a non linear space ? – Didier Dec 09 '20 at 15:03
  • normed linear space is a vector space equipped with some norm – Lucas Dec 09 '20 at 15:04
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    That is not my question. You are asking if $x+U$ is open in a general setting. I am pointing at the fact that $x+u$ may not even make sense in a general setting. Moreover, the proof in the normed case use the important fact that the norm is translation-invariant. – Didier Dec 09 '20 at 15:05
  • Okay, can you give me one such example where the sum doesn't make sense? – Lucas Dec 09 '20 at 15:06
  • He's saying that $x+u$ doesn't even make sense if you do not have an operation. The question is pretty much meaningless as currently stated. – Ivo Terek Dec 09 '20 at 15:06
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    okay got it, in discrete metric space there is no such operation so the sum is not defined – Lucas Dec 09 '20 at 15:07
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    Not only in a discrete space. In a general topological space (even non-discrete) there is no use of any addition. Here is a counter example for you: $\mathbb{R}$ with the topology ${ \varnothing,\mathbb{R}, {0},\mathbb{R}^*}$. ${0}$ is open but $1 + {0}$ is not, even if $\mathbb{R}$ is a vector space. – Didier Dec 09 '20 at 15:09
  • I'm sorry, I just realized that my counter-example above is not a metric space. But one can still find counter-example with "highly-non-translation-invariant" metrics. – Didier Dec 09 '20 at 15:18

1 Answers1

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If $X$ is a topological vector space, then by definition the sum operation $+$ is continuous. Since $U$ is open, and $f(y)= y+x$ is a homeomorphism (it is continuous and its inverse $g(z)=z-x$ is also continuous) then also $f(U)=U+x$ is open.

Bargabbiati
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  • Hi. The question is not about this proof, but about "if this still true in a non-normed space". – Didier Dec 09 '20 at 15:11
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    Topological vector spaces are non-normed spaces in general. – Bargabbiati Dec 09 '20 at 15:12
  • You are right. It seems OP was thinking about "general metric spaces" for which the question did not make sense (see the discussion in the comment sections). In the setting of topological vector spaces you are obviously right. – Didier Dec 09 '20 at 15:15
  • I haven't read anything about homeomorphism yet, so I couldn't understand. but thanks for answering. Will revisit this answer soon. :) – Lucas Dec 09 '20 at 15:17
  • A homeomorphism is simply a continuous bijective map whose inverse is continuous. They have the property of mapping open sets in open sets (try to check!). – Bargabbiati Dec 09 '20 at 15:21
  • The same logic works in the topological group setting too. – user854214 Dec 09 '20 at 16:14