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I am going through the basics of topology, in order to deal with topological vector space. I haven't taken any course of topology so I have some fundamental questions (I 've seen only some topological notion and definitions in other courses) Well, let $X$ be a topological vector space and $x\in\ X$

  1. A neighbourhood $V$ of a point $x$ is always consider to be an open set and why? I know that neighbourhoods are defined as sets containing an open set that contains $x$ but I can't figure out how a neighbourhood is an open set by itself. Same question about the basis of neighbourhoods of $x$

  2. We know that if $A\subseteq X$ an open set then $co(A)$ is also an open set (where $co(A)$ is the convex case of $A$) How can someone conclude from the above that $co(A)$ is a neighbourhood of $0$ ?

  3. Is the following right or have I misunderstood something ?

    $"$ Let $a\in\ K $ where $K$=($\Bbb R$ or $\Bbb C$) . Then we have $ax$ vector and $V$ a neighbourhood of $ax$ . There is neighbourhood $U$ of $x$ and $r \gt 0$ such as if $b\in\ K$ : $|b-a|\lt r$ and $y\in\ U$ such as $bx\in\ V$, more specific $bU\subseteq V$ $"$

    If this sentence is true , could somebody explain it to me with simple words?

I would appreciate any help! Thanks in advance!!

kaithkolesidou
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    Some authors require that a neighborhood $N$ of a point $p$ to be an open set, others use a different definition where $N$ need not be open, but there must be some open $U\subset N$ with $p\in U$. It just depends on context. – Math1000 Mar 29 '16 at 23:40

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Ad 1: A neighborhood can be defined to be always open, but as far as I know a neighborhood $N \subset X$ of $x \in X$ is usualy defined as a subset of $X$ containing some $U$ open with $x \in U$. $N$ doesn't need to be open. For instance $[0,1]$ is a neighborhood of the point $x = \frac{1}{2}$ as $ x\in (0,1) \subset[0,1]$.

Ad 2: For this to hold you need $0 \in co(A)$. In the general setting you gave it's not necessarily true:

Consider the open ball $B_1(3,3) = \{y \in \mathbb R^2 \mid \Vert y - (3,3) \Vert_2 < 1\}$. Obviously the origin is not in $B_1(3,3)$, but $co(B_1(3,3))=B_1(3,3)$ as the ball is convex itself.

So now assume $0 \in A$ for a moment: Since $A$ is open you know that $co(A)$ is open and hence $co(A)$ is a neighborhood of $0$ (as it contains itself and is an open set).

Ad 3: I have trouble understanding what it means. Is $K$ supposed to be the underlying field of the vector space? What are your assumptions, what the implications?

noctusraid
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  • I know the same definition about neighbourhood of a point $x$ but in every proof I see, neighbourhoods are considered to be open sets... – kaithkolesidou Mar 29 '16 at 23:00
  • It's convention. Probably they are using the definition which automatically assumes a neighborhood to be open. I rarely came across theorems or statements though where the neighborhoods weren't open. – noctusraid Mar 29 '16 at 23:07
  • For the 2nd question you use the definition of a neighbourhood as an open set, right? How do I know that $0\in\ co(A)$ ? – kaithkolesidou Mar 29 '16 at 23:15
  • You need some more structure on $A$ to say that, that's why I wrote "for this to hold you need $0 \in co(A)$". – noctusraid Mar 29 '16 at 23:19
  • For the 3rd question , this is what exactly the professor wrote on the board. I had also trouble of finding out the assumptions or the implications. When I asked him about it he told me this " if $V$ a neighbourhood of $ax$ then there is $r\gt 0$ : $|b-a|\lt r$ such as $ax-bx\in\ V$ " Is this more clear now..? – kaithkolesidou Mar 29 '16 at 23:27
  • The only thing we know about $A$ is that $ A\subseteq X$ where $X$ is a topological space.. Thank you very much for your answers and your time by the way!! – kaithkolesidou Mar 29 '16 at 23:30
  • I elaborated my answer on my laptop but my internet crashed and I'm on my phone right now. I'll post it tomorrow morning and have a look at 3 again. You're welcome!! – noctusraid Mar 29 '16 at 23:48
  • It's ok.. I 've got to attend a lesson early in the morning so if you post it tomorrow, it 'll be great! Thanks again! You helped me a lot..!! – kaithkolesidou Mar 30 '16 at 00:05