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Sep 3, 2018 at 2:25 audit Low quality posts
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Aug 17, 2018 at 7:49 history edited mathcounterexamples.net CC BY-SA 4.0
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Aug 13, 2018 at 16:47 comment added mathcounterexamples.net Also when you say *Apparently, equivalent norms induce equivalent topologies... *. It is much more than that. Equivalent norms induce the SAME topology. And by the way I don’t know what means equivalent topologies.
Aug 13, 2018 at 16:43 comment added mathcounterexamples.net For sure. That would be horribly confusing to define the differentiability using the limit based on the topology induced by the norm and to speak of opens of another topology for continuity or other topics!
Aug 13, 2018 at 13:28 comment added Our @mathcounterexamples.net Apparently, equivalent norms induces equivalent topologies on the underlying set, but now the question is this; for differentiability, when we talk about an open set of a normed space, are we referring simplicity the open sets that are coming from the metric topology on that space induced by the norm ?
Aug 13, 2018 at 12:01 comment added Our @mathcounterexamples.net by the way, it would be also nice if you just checkout the comments under Perturbative's answer.
Aug 13, 2018 at 11:53 comment added Our @mathcounterexamples.net I do not want to turn this into an XYZ question, but rather just want to make sure that I have indeed understood the given answer; When you say that all the norm's on $\mathbb{R}^n$ are equivalent, do you mean that all induces the same topology on $\mathbb{R}^n$, or something else ? (I haven't taken a Mathematical Analysis course, so I do not have general knowledge on normed / metric spaces, but still I have the basics).
Aug 13, 2018 at 10:30 comment added Our @mathcounterexamples.net I wish you asked me to clarify myself. I mean I'm not a native English speaker, and even in my native language, I'm not good at explaining what I mean if I'm not doing pure math, so please ask or just leave a comment about the vague points.
Aug 13, 2018 at 10:20 comment added mathcounterexamples.net That would be great. I already tried to answer some of your other questions that were lacking precision!
Aug 13, 2018 at 10:19 comment added Our I'm going to try to edit to question to make it clear what I'm asking.
Aug 13, 2018 at 10:18 comment added Our @mathcounterexamples.net I don't want to disrespect you, nor I want to undermine the effort you have made you answer and help me; however, the answer (the others also, except Jose's) does not relevant to what I'm actually asking.
Aug 13, 2018 at 8:55 history edited mathcounterexamples.net CC BY-SA 4.0
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Aug 13, 2018 at 8:49 comment added mathcounterexamples.net @GiuseppeNegro That doesn’t make any difference. Relative topology is defined for subsets, not only subspaces.
Aug 13, 2018 at 8:46 comment added Giuseppe Negro Subsets of $\mathbb R^n$, actually. The OP uses the word "subspace" but the reference is to arbitrary subsets.
Aug 13, 2018 at 8:43 comment added mathcounterexamples.net @GiuseppeNegro But the precise OP question doesn’t deal with manifolds but with subspaces.
Aug 13, 2018 at 8:41 comment added Giuseppe Negro The question concerns differentiable manifolds. In a differentiable manifold, there might co-exist more differential structures on the same topology, in contrast to what happens in normed spaces.
Aug 13, 2018 at 8:17 comment added Our yes, but I haven't seen a definition of differentiability that considers $[0,1]$ as the bigger space - let $U$ be open in $[0,1]$ - so what is the reason for that ? what is special about $\mathbb{R}^n$ ? why are we never considering differentiability in a subspace of $\mathbb{R}^n $ ?
Aug 13, 2018 at 8:13 comment added mathcounterexamples.net That is not different for subspaces. The relative topologies are induced by topologies defined by equivalent norms. Therefore, there is only one relative topology.
Aug 13, 2018 at 8:08 comment added Our what about in a subspace of $\mathbb{R}^n $ ?
Aug 13, 2018 at 8:07 history answered mathcounterexamples.net CC BY-SA 4.0