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August 31st, 2017, 08:40 AM   #51
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Quote:
 Originally Posted by cjem In Sutherland page 94, "A neighbourhood of a point $x$ in a space $X$ is a subset $N$ of $X$ which contains an open subset of $X$ containing $x$." There's actually a remark after that says, "Some writers insist that a neighbourhood itself should be open in X." My lectures followed this definition. On page 39 of the Cambridge notes, "Let $(X, \tau )$ be a topological space. If $x \in X$, we say that $N$ is a neighbourhood of $x$ if we can find $U \in τ$ with $x \in U \subseteq N$."
(Relatively open sets in the current context confuses me too, so let's leave it ou)

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- non-metric. 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
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Quote:
 Originally Posted by zylo ...
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
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R,
Quote:
 Originally Posted by cjem 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.
I have no quibble with any definition which makes sense.

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
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Quote:
 Originally Posted by zylo 1) An open set is defined by neighborhoods. So "neighborhoods", as you define them, include neighborhoods.
To avoid this issue of circular definitions, you could equivalently define a set $A \subseteq \mathbb{R}$ to be open if for all $a \in A$, there exists $\epsilon > 0$ such that $(a - \epsilon, a + \epsilon) \subseteq A$.

Quote:
 Originally Posted by zylo 2) So is (1.2)U(3,4) a "neighborhood" of .5?
It is a neighborhood (and an open neighborhood) of $\frac{1}{2}$ according to the definitions I've listed.

Quote:
 Originally Posted by zylo 3) Your definition says any set which contains a neighborhood of x is a "neighborhood" of x.
It says any set which contains an open neighborhood of x is a "neighborhood" of x, but this is equivalent to what you've said.

Quote:
 Originally Posted by zylo ...
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
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Quote:
 Originally Posted by zylo 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).
Anyone can grasp my definition. It is absolutely basic and all-encompassing.
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
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Quote:
 Originally Posted by zylo Anyone can grasp my definition. It is absolutely basic and all-encompassing. The Union and Intersection conditions fall under "specific collection."
I was fine with the part you've quoted- it's mainly the stuff about distinguishing open sets and "open sets" that I'm not really getting.

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 $(S,s) = \{S, \varnothing, \{1\}, \{3\}, \{7\}, \{1,7\},\{1,3\},\{3,7\}\}$.

Quote:
 Originally Posted by zylo But I really appreciate all the lucid comments from an expert, many thanks.
No worries! I'm not really an expert, but thanks for the compliment!

September 1st, 2017, 10:46 AM   #57
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Quote:
 Originally Posted by cjem I was fine with the part you've quoted- it's mainly the stuff about distinguishing open sets and "open sets" that I'm not really getting. Finally, the collection $\{S, \varnothing, \{1\}, \{3\}, \{7\}, \{1,7\},\{1,3\},\{3,7\}\}$ is the topology $s$ on $S$, not the topological space $(S,s)$. No worries! I'm not really an expert, but thanks for the compliment!
Quote:
 Originally Posted by zylo 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.

(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
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Quote:
 Originally Posted by Maschke You have all the information you need. Start by writing down exactly what you need to show.
Hello. Sorry for the very late reply since I have an exam, I was not able to actively participate here. Here is my proof ( kind of ).

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
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Quote:
 Originally Posted by Lalitha183 $\displaystyle \forall \epsilon >0$ $\displaystyle \exists$ $\displaystyle 0 \epsilon [0,1)$
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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