August 31st, 2017, 08:40 AM  #51  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88  Quote:
If we are talking about metric spaces, and the usual definition of open, then adding open to Sutherland is redundant, as I pointed out earlier. "Cambridge notes" doesn't mention "open sets," but there is the confusion of U and N, which are both neighborhoods nonmetric. U is just a set containing x. But what does that mean? I note the same confusion with Sutherland, where therev is a "neighborhood of x in the sense of Rudin, and the "Neighborhood" N. Are Sutherland and "Cambridge" talking about topological METRIC spaces? I note the wiki comment: "metric spaces are a class of Topological spaces." Se are we (you) also talking about Topology without a metric, in which case what does an open set mean? You dismissed Rudin as irrelevant because he used distance (a metric). I note the OP definition IN R is clear. As for the rest of the thread, general case of Topological Space, I find it a confusion of definitions.  
August 31st, 2017, 09:16 AM  #52 
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  As you're unfamiliar with topological spaces, it's probably best if we ignore them for this discussion and restrict ourselves to $\mathbb{R}$ (using the definition of open that you are familiar with). In this setting, what all the definitions I've listed say is: "a neighborhood of a point $x \in \mathbb{R}$ is a set $N \subseteq \mathbb{R}$ containing an open set containing $x$". While this sounds a bit awkward, hopefully you can see that it's equivalent to saying: "a neighborhood of a point $x \in \mathbb{R}$ is a set $N \subseteq \mathbb{R}$ such that $(x\epsilon,x+\epsilon) \subseteq N$ for some $\epsilon > 0$." Notice that, for example, $[0,1]$ is not an open set, but it is a neighborhood of $\frac{1}{2}$ according to this definition: $[0,1]$ contains $(0,1)$ which is an open set containing $\frac{1}{2}$. Anyway, the point of me saying all this is just to show that some prolific authors (like Rudin) require neighborhoods to be open in their definition, while others (Hatcher, Sutherland, etc.) don't. Last edited by cjem; August 31st, 2017 at 09:53 AM. 
September 1st, 2017, 07:18 AM  #53  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88 
R, Quote:
1) An open set is defined by neighborhoods. So "neighborhoods", as you define them, include neighborhoods. 2) So is (1.2)U(3,4) a "neighborhood" of .5? 3) Your definition says any set which contains a neighborhood of x is a "neighborhood" of x. (I can see this for a connected set, and I seem to recall seeing this as somebody's classic definition, in which case I can see the attempt at extrapolation (generalization)). Basic concept (building blocks) of Topology: A set and all it's subsets. "X" Topology: A Set and a specific collection of subsets known as "open subsets." (S,s). Example of an "X" Topology: (S,s)={S,1,3,7,{1,7},{1,3},{3,7}} What are the open subsets of S? You say there aren't any? You obviously don't know anything about Topology, ie, you are not aware of my definition. Note the confusion caused by open subset for "open subset." Another example of an "X" Topology: S = {1,3,7} (S,s) = {S,{1,7},{1,3},{3,7}} Another example of an X Topology: S=R (S,s)={R,(1,2),[5,7],(1,2)U[5,7)]} What are the open subsets? What are the "open subsets?" Last edited by zylo; September 1st, 2017 at 07:23 AM.  
September 1st, 2017, 07:55 AM  #54  
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  Quote:
It is a neighborhood (and an open neighborhood) of $\frac{1}{2}$ according to the definitions I've listed. Quote:
I'm afraid I can't really understand what you're trying to say about topology  your phrasing seems a bit off. Given any set $X$, you can declare any subsets of $X$ to be open so long as certain conditions are satisfied: $X$ and the empty set are open, any countable union of open sets is open, and any finite intersection open sets is open. This collection of open sets is then called a "topology" on $X$. So, for example, if $X = \mathbb{R}$, one topology would consist of just $\mathbb{R}$ and the empty set  this is called the indiscrete/trivial topology on $\mathbb{R}$. Another topology would consist of the sets that are open with respect to the standard metric on $\mathbb{R}$  it's an instructive exercise to show that this actually is a topology by checking against the above conditions. Similarly, given any metric space $(X,d)$, you can show that the collection of all subsets of $X$ that are open with respect to $d$ is a topology on $X$. Last edited by cjem; September 1st, 2017 at 08:34 AM.  
September 1st, 2017, 09:39 AM  #55  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88  Quote:
The Union and Intersection conditions fall under "specific collection." But I really appreciate all the lucid comments from an expert, many thanks.  
September 1st, 2017, 10:17 AM  #56  
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  Quote:
Just some nitpicks: $\{S,1,3,7,\{1,7\},\{1,3\},\{3,7\}\}$ is not a topology on $S = \{1,3,7\}$ as it does not contain the empty set (and you should have set brackets around the $1$, $3$ and $7$). But it's fine if you instead say $\{S, \varnothing, \{1\}, \{3\}, \{7\}, \{1,7\},\{1,3\},\{3,7\}\}$. Finally, the collection $\{S, \varnothing, \{1\}, \{3\}, \{7\}, \{1,7\},\{1,3\},\{3,7\}\}$ is the topology $s$ on $S$, not the topological space $(S,s)$. So you should be saying $s = \{S, \varnothing, \{1\}, \{3\}, \{7\}, \{1,7\},\{1,3\},\{3,7\}\}$ rather than $(S,s) = \{S, \varnothing, \{1\}, \{3\}, \{7\}, \{1,7\},\{1,3\},\{3,7\}\}$. No worries! I'm not really an expert, but thanks for the compliment!  
September 1st, 2017, 10:46 AM  #57  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88  Quote:
Quote:
But it has grown on me. And it's so easy to recall and reference that you shouldn't get confused by "open set" and open set. (S,s). It's my definition. But not to worry. My definitions will last till the next page, or sooner.  
September 1st, 2017, 11:32 AM  #58 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88 
Ref previous post: It would have been fair not to quote me out of context in this case. The second sentence is important and not that difficult to add. So to correct that lapse, this is the quote with relative context: "Basic concept (building blocks) of Topology: A set and all it's subsets. "X" Topology: A Set and a specific collection of subsets known as "open subsets." (S,s). Frankly, I meant that definition to be sarcastic to show the confusion that can arise from redefining or generalizing existing terms without specifically pointing out that you are doing so, as illustrated in my examples. But it has grown on me. And it's so easy to recall and reference that you shouldn't get confused by "open set" and open set." 
September 18th, 2017, 11:09 AM  #59  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
We have the set $\displaystyle (0,1)\cup $ {$\displaystyle 1,2,....,n$} (i)For $\displaystyle n=1$ : $\displaystyle (0,1) \cup${$1$} $\displaystyle \forall \epsilon >0$ $\displaystyle \exists$ $\displaystyle 0 \epsilon [0,1)$ such that $\displaystyle (0\epsilon, 0+\epsilon)\not\subset[0,1)$ (or) $\displaystyle \forall \epsilon >0$ $\displaystyle \exists$ $\displaystyle 1 \epsilon (0,1]$ such that $\displaystyle (1\epsilon, 1+\epsilon)\not\subset(0,1]$ (ii)For $\displaystyle n=2$ : $\displaystyle (0,1) \cup${$2$} $\displaystyle \forall \epsilon >0$ $\displaystyle \exists$ $\displaystyle 0 \epsilon [0,1]$ such that $\displaystyle (0\epsilon, 0+\epsilon)\not\subset[0,1]$ and $\displaystyle \forall \epsilon >0$ $\displaystyle \exists$ $\displaystyle 1 \epsilon [0,1]$ such that $\displaystyle (1\epsilon, 1+\epsilon)\not\subset[0,1]$ For $\displaystyle n>2$ : $\displaystyle \forall \epsilon >0$ there exists some $x$ in the $\displaystyle \epsilon$interval except $0$ & $1$ (from (ii)) such that $\displaystyle (x\epsilon, x+\epsilon) \not\subset [0,1]$ As you have included in the previous post, I'm taking $\displaystyle (0,1] \cup$ {$2$} to state that $n>2$ does not form an interval. I have understood the thing, but I don't know whether it is the correct way to represent it. Correct it if it has any errors! Thank you  
September 18th, 2017, 11:17 AM  #60 
Senior Member Joined: Aug 2012 Posts: 1,514 Thanks: 364  This doesn't parse. Even allowing $\epsilon$ to mean $\in$ it still doesn't make sense. How did $(0,1)$ suddenly become $[0,1)$? Clarify please. Also please state what it is you're proving. This question has multiple parts and I don't know which part you are addressing.
Last edited by Maschke; September 18th, 2017 at 11:22 AM. 

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