February 27th, 2017, 07:15 PM  #31  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
Sure today I will go through these terms and will provide you the same so that you can rectify if anything goes wrong.  
February 27th, 2017, 07:22 PM  #32 
Senior Member Joined: Aug 2012 Posts: 1,922 Thanks: 534  Ok sounds like a plan. I promise you that once we nail down these concepts in the real numbers, you will laugh at general topology. It's very simple once you have the concepts clear in your mind in the real numbers.

February 27th, 2017, 09:27 PM  #33  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
What is an open set? Is it a set which has infinitely many points between the intervel a and b and the endpoints a and b are not included ? when I refer to a definition where $\epsilon > 0$ and a real number $x ,y \epsilon R$ .If the condition $dist(x,y) < \epsilon$ holds is an openset. What it means ? Let us take (1,4) and the dist(1,4) = 3. let us take as $\epsilon = 5$ which satisfying the condition $\epsilon >0$ So, Is (1,4) is an open set ? Can you show me the openset in a pictorial way ? That would be more helpful  
February 28th, 2017, 05:42 AM  #34  
Senior Member Joined: Aug 2012 Posts: 1,922 Thanks: 534  Quote:
I started writing up an explanation that will clear all this up for you. I'm traveling today and hopefully I'll be able to post something later tonight or tomorrow. Very briefly, if we define the epsilonball around a point $x$ as $\mathcal B_\epsilon(x) = \{y \in \mathbb R : d(x,y) < \epsilon\}$ where $d$ is the distance function $d(x,y) = x  y$ then * If $X$ is a subset of reals, a point $x \in X$ is called an interior point if there's some epsilon ball around $x$ entirely contained in $X$; and * The set $X$ is called open if all its points are interior points. I haven't got time to write more but I have a writeup that starts from that and goes all the way through the definition of a topological space and I'll have that by tomorrow. From these definitions see if you can prove that in the unit interval $[0,1]$ the point $\frac{1}{2}$ is an interior point and the point $1$ is not. Therefore $[0,1]$ is not an open set. Then prove that $(0,1)$ is an open set. Then see if you can prove that any open interval $(a,b)$ is an open set, and that the union of two open sets is open, and then that the arbitrarily union of open sets is open. I'll have those proofs written out for you tomorrow. Possibly with pictures if I can find time to do that. Meanwhile this Wiki article has a great picture of an interior point. https://en.wikipedia.org/wiki/Interior_(topology) ps  Also see if you can prove from the definitions I gave that the empty set and the reals are open subsets of the reals. And see if you can prove that finite intersections of open sets are open. Start with the intersection of two open sets. Last edited by Maschke; February 28th, 2017 at 06:35 AM.  
March 1st, 2017, 08:54 AM  #35 
Senior Member Joined: Aug 2012 Posts: 1,922 Thanks: 534 
I enjoyed writing this. I hope you find value in it. In my opinion to study topology you should spend some time with this material. The intuitions for basic topology all come from the real numbers. Intro This is an overview of the usual topology on the reals, along with proofs that the open sets of reals satisfy the definition of an abstract topology. After all the preliminaries I get to your original question and prove that a subet of a topological space is open if and only if all its points are interior points. I was wrong earlier, the definition of interior point in your video is correct. In $a \in U_a \subset A$ the inclusion need not be proper. That became clear when I couldn't get the proof to work without it. Prerequisites We assume the basic language and terminology of set theory up to and including arbitrary unions and intersections, countable and uncountable sets, and the use of the universal and existential quantifiers $\forall$ and $\exists$. We assume the existence and basic arithmetic and order properties of the set $\mathbb R$ of real numbers. We have the notion of open and closed intervals of real numbers, $(a,b)$ and $[a,b]$. At this point we have not yet proved that an open interval is an open set, so the word doesn't mean anything yet. We have a map $\lvert \cdot \vert : \mathbb R \to \mathbb R_{\geq 0}$ that maps the real number $x$ to its absolute value $\lvert x \vert$. The absolute value has two important properties of interest to us: * (Symmetry): $\forall x, \lvert x \vert$ is $0$ if and only if $x = 0$. * (Triangle inequality): $\forall, x, y \in \mathbb R, \lvert x + y \vert \leq \lvert x \vert + \lvert y \vert$. Given two real numbers $x$ and $y$ we define their distance $d(x,y) = \lvert x  y \lvert$. This notation is useful since it generalizes to metric spaces without any additional work. The properties of $d$ that we care about are ): * (Nonnegativity): $\forall x,y \in \mathbb R, d(x,y) \geq 0$. * $\forall x,y \in \mathbb R, d(x, y) = 0$ if and only if $x = y$. * (Symmetry): $\forall x,y \in \mathbb R, \ d(x,y) = d(y, x)$. * (Triangle inequality): $\forall x,y,z \in \mathbb R, d(x,z) \leq d(x,y) + d(y,z)$. Each of these can be proved directly from the properties of the absolute value. Those are the prereqs for the rest of the article. If any of this is not clear please ask. That goes for anything in here. Needless to say there may be errors so if something seems off let me know. Open sets of reals If $x, \epsilon \in \mathbb R$ and $\epsilon > 0 $, we define the open ball of radius $\epsilon$ around $x$ as $\mathcal B_\epsilon(x) = \{y \in \mathbb R : d(x, y) < \epsilon\}$. This is the key concept in all of topology, really. A little cloud of points around $x$ that is "missing its boundary," a phrase that's made precise in topology. Conceptually, an open ball is a little dottedline circle around a point. In ther real numbers, an open ball is an open interval. Ok now we have the reals and $\epsilon$balls. From these we will build an empire. If $A \subset \mathbb R$ and $a \in A$, we say that $a$ is an interior point if $\exists \epsilon > 0 : \mathcal B_\epsilon(a) \subset A$. In other words we can draw a little $\epsilon$ ball around $a$ that stays entirely within $A$. Example: In the closed unit interval $I = [0,1]$ the point $a = \frac{1}{2}$ is an interior point and the point $b = 1$ is not. (You should make sure you can prove those from the definitions). Now if $A \subset \mathbb R$ then we say $A$ is an open set if all its points are interior points. In other words in the same way that an open ball is a little circle without its boundary (which of course we have not formally defined yet) then an open set is a lumpy random shape or blob that is also missing its boundary. Now the blob above is all one piece. But an open set can have multiple parts. It could look like three blobs, or even infinitely many blobs. For exampethe real numbers minus the integers are an open set. ... ()3()4()5()6()7()... We can notate this set as $\mathbb R \setminus \mathbb Z = \cup_{n \in \mathbb Z} (n, n+1)$. This is an open set that is a countable union of open intervals. It turns out that every open set of reals is a countable (or finite) union of open intervals, but that proof is a little beyond us at the moment. The empty set is open I claim the empty set is open, using the following vacuous argument. For the open set to fail to be open, we'd have to find some point $x \in \emptyset$ such that there is no epsilonball about $x$ contained in the empty set. But we can never do that. Why? Because there's no $x \in \emptyset$ in the first place. So $\emptyset$ is open. This type of argument usually startles people the first time but after a while you get used to it. All the pink elephants in the empty set can fly. Also true. If this isn't completely clear to you, you're not the only one. But it's how logic works and when we do math we agree to think like this. The reals are open in the reals The real numbers are an open subset of themselves. Given any real number $x$, we can take $\epsilon = 1$ say and there's an open interval around each real entirely contained in the reals. If we only knew that an open interval is an open set, we'd be done. Proof left to the reader. Why did I use the awkward phrase "the reals are open in the reals?" Because being an open set of a topological space is a relative property. If I pick up the real line and put it in the complex plane, it's no longer an open set. You can see that if you put an epsilon ball around any real number in the complex plane, that epsilon ball always contains some nonreal numbers. A given set may be open as a subset of one topological space but not another. An arbitrary union of open sets is open Theorem: An arbitrary (finite, countably infinite, or uncountable) union of open sets is open. Proof: Let $\mathscr A$ be an index set, and for each $\alpha \in \mathscr A$ let $U_\alpha$ be an open set. Let $U = \cup_{\alpha \in \mathscr A} U_\alpha$ be their union. If $u \in U$ then by the definition of union, there exists a specific $\alpha \in \mathscr A$ such that $u \in U_\alpha$. Since $U_\alpha$ is open there is some $\epsilon > 0$ such that $\mathcal B_\epsilon(u) \subset U_\alpha \subset U$ so $U$ is open. $\square$ A finite intersection of open sets is open How about intersections? It seems clear that if we have two overlapping open sets, their intersection is open. Would this work for infinite intersections? An example shows that it would not. Ex: For $n \in \mathbb N$ let $I_n = (\frac{1}{n}, \frac{1}{n})$. What is the intersection $\cap_{n \in \mathbb N} I_n$? It's the singleton set $\{0\}$, because $0 \in I_n$ for every $n$. But $\{0\}$ is not an open set. How do we know this? For any epsilon you could never fit an entire epsilonball inside $\{0\}$. We've just proved that in the reals with the usual topology, singletons are never open sets. So if infinite intersections of open sets aren't necessarily open, the next best thing we can prove is: Theorem: A finite intersection of open sets is open. Pf: Let $U_k$ be open for each $k$ in $\{1, 2, \dots, n)$ for some positive integer $n$. Let $u \in U = \cap_n U_k$. Then for each $k$ there is some $\epsilon_k > 0$ with $\mathcal B_{\epsilon_k}(u) \in U$. Now let $\epsilon = min(\epsilon_k)$. Note that the minimum exists exactly because we only have finitely many epsilons. That's the trick behind this proof. Then for each $k$, $\mathcal B_\epsilon(u) \subset \mathcal B_{\epsilon_k}(u)$ and therefore $\mathcal B_\epsilon(u) \subset U$ and $U$ is open. $\square$ Euclidean spaces Before getting to topological spaces I just want to mention that if we are in $n$dimensional Euclidean space, $\mathbb R^n = \{(x_1, x_2, \dots, x_n) : x_i \in \mathbb R\}$ then if $x = (x_i)_n$ we define the usual Euclidean norm $\lVert x \rVert = \sqrt{\sum_n x_i^2}$ and this induces the metric $d(x,y) = \lVert x  y \rVert$. When we study the topology of the line we are also getting a good start on the topology of Euclidean $n$space. Metric spaces Everything we've said so far works for metric spaces in general, which are just sets with a distance function. But metric spaces can be a lot different than Euclidean space. Example: The discrete metric is defined on the real numbers (or on any set in fact) as $d(x,y) = 0$ if $x = y$ and $d(x,y) = 1$ if $x \mathbb \neq y$. This distance function satisfies is a metric. We can therefore define open balls and open sets. In fact if $\{x\}$ is a singleton, then $\mathcal B_{\frac{1}{2}}(x)$ is an epsilonball entirely contained in $\{x\}$. In fact $\mathcal B_{\frac{1}{2}}(x) = \{x\}$. In other words every singleton is open. And since every set is the union of its singleton elements, every set is open. In this case the topology is the entire powerset of the original set. Topological spaces Now we are ready to define a topological space. Def: A topological space is a set $X$ along with a collection of subsets $\mathscr T \subset \mathscr P(X)$ (where $\mathscr P(X)$ is the powerset of $X$, the set of all the subsets of $X$) such that: * $\emptyset \in \mathscr T$; * $X \in \mathscr T$; * $\mathscr T$ is closed (also sometimes called stable) under arbitrary unions; and * $\mathscr T$ is closed under finite intersections. We denote a topological space as $(X, \mathscr T)$. If the topology is understood we will often refer to "the topological space $X$" and omit explicit mention of $\mathscr T$. Theorem: The real numbers with the open sets as defined earlier are a topological space. Pf: We walked through this above. $\square$ Example: The discrete topology on a set $X$ is the topology $\mathscr T = \mathscr P(X)$. In other words, every set is open. This is the exact same topology that we got from the discrete metric. So in this case we can define a topology and then find a metric that gives us exactly that topology. Example: The indiscrete topology on $X$ is the topology $\mathscr T = \{\emptyset, X\}$. The only open sets are the empty set and the whole set. What is interesting is that there is no metric whatsoever that will produce the indiscrete topology (needs proof of course). A topology that can be induced by a metric is called metrizable. The discrete topology is metrizable and the indiscrete topology is not. Interior points Now we are finally in a position to address the question you originally asked! If $(X, \mathscr T)$ is a topological space and $A \subset X$, a point $a \in A$ is called an interior point of $A$ if there exists an open set $U_a$ with $a \in U_a \subset A$. NOTE! When we write $a \in U_a \subset A$ it may be the case that $U_a = A$. I was wrong about this earlier. In fact if $A$ is already known to be an open set, then any point $a \in A$ is in some open set contained in $A$, namely $A$ itself. We've just proved one direction of: Theorem: A set $A$ in a topological space $X$ is open if and only if each of its points is an interior point. Proof: As we just noted, if $A$ is open and $a \in A$, then $a \in A \subset A$ satisfies the definition of an interior point. Since $a$ is arbitrary, every point of an open set is an interior point. To show the other direction, suppose that we have some set $A$, and each of its points is an interior point. We want to show that $A$ must be open. For each $a \in A$ let $U_a$ be an open set with $a \in U_a \subset A$. Since an arbitrary union of open sets is open, the union $U = \cup_A U_a$ is an open set. Now I claim that in fact $U = A$ which would show that A is indeed an open set. How do I prove $U = A$? I have to show (1) that each element of $U$ is in $A$ and (2) each element of $A$ is in $U$. (1) If $u \in U$ then $u \in U_a$ for some $a \in A$. Since $u \in U_a \subset A$, we have $u \in A$ so $U \subset A$. (2). Suppose $a \in A$. Since $a$ is an interior point (remember we're assuming all points of $A$ are interior) there's an open set $U_a$ with $a \in U_a$. But then $a$ is in the union of all the $U_a$'s, namely $U$. So $A \subset U$. Having proved set inclusions in both directions, we have $U = A$. $\square$ Last edited by Maschke; March 1st, 2017 at 09:18 AM. 
March 1st, 2017, 08:59 AM  #36 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 
Thanks for all your efforts ðŸ˜Š. Tomorrow morning i will surely start reading it.

April 5th, 2017, 12:55 PM  #37 
Newbie Joined: Apr 2017 From: Kolkata Posts: 2 Thanks: 1 
Look Lalitha, the phrase "interior point" has no meaning until you say "interior point of some particular set under some particular topology". By interior point $x$ of a set $A$ under some topology, we mean a point of the set $A$, such that there is an open subset $U$ of $A$ containing that point. That is $x \in U \subseteq A$. So if you mean to say that a set whose every point is an interior point of it is open, then you are correct.
Last edited by Kuldeep Guha Mazumder; April 5th, 2017 at 12:58 PM. 

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doubts, interior points, open set, topology 
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