I'm studying about manifolds. I'm an engineer and want to grasp the concept of manifold without too deep dive into rigorous mathematics. My question is whethere the "topologist's sine curve" defined as $S = \{(0, y) : y \in (-1,1)\} \cup \{(x, \sin (1/x)) : x \in (0, 2/\pi)\}$ can be a manifold. I tried to find some explnations in books and web. However, what I found are all about (path/local) connectedness. I'm not familiar with topology and not want to study it extensively. So I tried to understand the question with very minimal knowledge.  As far as I understand, to say that a set is a manifold, I need to set a topology (open subsets) on the set. As a subset of $\mathbb{R}^2$, every open set in $S$ should have a form $S \cap A$ for an open set $A \in \mathbb{R}^2$. Any interval in the vertical axis part of $S$, e.g. $I = \{(0, y) : 0 &lt; y &lt; 1/2\}$, cannot be an open set in $S$. It is because every open set in $\mathbb{R}^2$ containing $I$ also contains some other part of $S$ : there is no way to express $I$ in the form of $S \cap A$ where $A$ is open in $\mathbb{R}^2$. Then there is no homeomorphism from an open subset of $S$ containing $I$ to an open set in $\mathbb{R}$. It is because, if there is one, the preimage of an open interval in $\mathbb{R}$, to which $I$ is mapped, must be open but $I$ is not open. So there cannot be a coordinate chart for $I$ and $S$ is not a manifold. This is what I understand. My first question is whether or not this understanding is correct.  The second question is whether $S$ can be a manifold if I give a different topology. I cannot express precisly how to define the topology but intuitively the sine function part is homeomorphic to an open interval in the real line and the vertical part is also homeomorphic to an open interval in the real line, if I consider $S$ as a space of its own, not a subspace of $\mathbb{R}^2$. Is this true? If true, how can I express the topology precisley?    This question is not a duplicate of . I read it. However, the concept of connectedness confuses me. As I said in the first paragraph, I want to understand my question with minimal knowledge on topology such as just the definition of open sets, continuous functions and manifolds. Moreover the linked question does not concern my second question.
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I am trying to work out the details of following example from page 101 of Tu's An Introduction to Manifolds:     Example 9.3. Let $\Gamma$ be the graph of the function $f(x) =  \sin(1/x)$ on the interval $]0, 1[$, and let $S$ be the union of   $\Gamma$ and the open interval $I=\{(0,y)\in \mathbb{R^2} |−1&lt;y&lt;1\}$.      The subset $S$ of $\mathbb{R^2}$ is not a regular submanifold for the   following reason: if $p$ is in the interval $I$, then there is no   adapted chart containing $p$, since any sufficiently small   neighborhood $U$ of $p$ in $\mathbb{R^2}$ intersects $S$ in infinitely   many components.   Using Tu's definitions, I need to show that given a neighborhood $U$ of $p$, there exists no homeomorphism $\phi$ from $U$ onto an open set of $V \subset\mathbb{R^2}$ with the property that $U \cap S$ is the pre-image with respect to $\phi$ of the $x$ or $y$ axes intersected with $V$.   I am not sure where the fact that $U \cap S$ has infinitely many components comes into play. I tried to derive a contradiction using this fact; but even if restrict my attention to connected neighborhoods $U$ of $p$, the intersection of the connected set $\phi(U)$ with the $x$ or $y$ axes might have infinitely many components (I think), so there's no contradiction with the fact that homeomorphism preserves components.  I would appreciate any help!
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It's well known that, in $\mathbb{R}^n$:     (1) Open and Connected $\Rightarrow$ Path-connected   The proof essentially goes through the fact that (2) Every path-connected component will be open. Using this fact, we arrive at a contradiction if we suppose there are more than one path-connected component.  Well, trying to see where (1) would keep being valid, I arrived at the following:     If $X$ is a locally convex topological vector space, then (1) is valid.   But, to prove this, I proved (2). But for that, you use the (rather strong imo) convexity of a local base (to essentially repeat the argument for $\mathbb{R}^n$). But I'm not satisfied with this, as I think that we are using a lot of strong conditions (using the existence of a "line segment" to prove the existence of a "curve" seems very bazooka-like to me).  Anyway, given the previous considerations, my questions are:  Are there more general examples of spaces where (1) holds? Is there a characterization of spaces that satisfy this property?  And bonus question, since (2) implies (1) if the set is open.:  Are there more general examples of spaces where open sets satisfy (2)? Is there a characterization spaces that satisfy this property?
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Calculate the number $o(\mathbb{R})$ of open subsets of the real line.  I know that the answer is $\mathfrak{c}$ but I don't know how my lecturer got this.  I am doing an introductory topology course, so I know a few topological concepts. Such as that the set of all open intervals forms a base for the real numbers. Is this somehow relevant in showing that  $o(\mathbb{R}) = \mathfrak{c}$?
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Let $X$ be a subset of $\mathbb{R^n}$. The topology on $X$ is induced by the topology of $\mathbb{R^n}$. If there is an homeomorphism from $X$ onto $\mathbb{R}^n$, is it true that $X$ is open in $\mathbb{R}^n$ ? I think yes, but how can it be shown? (of course, $X$ is open in $X$).  If $x \in X$, $H_{n-1}(\mathbb{R}^n-\{f(x)\})= \mathbb{Z}$ so $H_{n-1}(X-\{x\})= \mathbb{Z}$, but I don't know how to continue.  Thanks in advance.
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Let $f(x)=\sin(1/x)$. If I'm correct then the boundary of $S=\{(x,f(x)):x\in\mathbb{R}\} \subset \mathbb{R}^2$ is $\{(x,f(x)):x \text{ in domain of } f(x)\}\cup\{(0,0)\}$, since for any $\varepsilon &gt; 0$ and an open ball $B_\varepsilon(0)$, $B_\varepsilon(0)\cap S\ne \emptyset \ne B_\varepsilon(0)\cap [S]^c$. [I have some doubts about this part though].  For the interior of $S$, it is the empty set, since there does not exist an $\varepsilon &gt; 0$ such that $B_\varepsilon((x,f(x)))\subset S$.  Thus, the closure of $S$ is $S\cup\{(0,0)\}$.  Please let me know if you think this is about right.
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Proposition : The only sets in $R^n$ which are both open and closed are the empty set and $R^n$ itself.  I came up with the proof of this claim and I'd like to know if my proof is correct.  Proof:  Let $S$ be the nonempty proper subset of $R^n$ which is both open and closed. Then $T = R^n-S$ is also a nonempty proper subset of $R^n$ which is both open and closed. Now take a point $s_0$ in $S$ and $t_0$ in $T$. Then we can consider a closed box $J_1$ which contains both $s_0$ and $t_0$.  The n-dimensional interval $I_1^k$ is defined by the inequalities   $s_k\le x_k \le t_k$ (k=1, 2, $...$, n ), where each $s_k$ and $t_k$ are the k-th coordinates of $s_0$ and $t_0$. Without loss of generality, assume that for all k, $s_k \le t_k$.  Here $J_1$ denotes the catesian product   $J_1$ = $I_1^1$ x  $\cdots$ x $I_n^1$   Now, bisect all of the n 1-dimensional intervals and the point $\frac {s_0+t_0}{2}$ will lie in the center of the bisected box. This point belongs to either $S$ or $T$. Say it belongs to $S$ then, rename it $s_1$ and the original $t_0$ to be $t_1$. Now again consider a box that has both points as vertices like the previous case and repeat the process.   Writing $I_k^m$ = [$s_k^m$,$t_k^m$],   we have  $t_k^m$ - $s_k^m$ = $\frac {t_k-s_k}{2^{m-2}}$  Then for each fixed k, the sup of all left endpoints must therefore be equal to the inf of all right endpoints, and their common value we donote by $b_k$ (actually we know this by Bolzano-Weierstrass).   Now we assert that b=($b_1$,$...$,$b_n$) is a limit point of $S$ and $T$. To see this, take any n-ball B(b; $r$). The point b of course belongs to each of the boxes $J_1$, $J_2$, $...$ constructed above, and when $m$ is such that $max_k\frac {t_k-s_k}{2^{m-2}}$ &lt; r/2, this neighborhood will include $J_m$ and since $J_m$ contains both points of $S$ and $T$, b is a limit point of both sets. But since both sets are closed, they both contain b, which is a contradiction. Hence, the proposition is true.
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This has been asked about here: , and  said      Personally I'd say $M$ wasn't a valid manifold with boundary because the $\{2\}$ doesn't have a neighborhood with any structure like an open ball/half-ball.    This is actually an exercise from An Introduction to Manifolds by Loring W. Tu and is not mentioned in an .        I have spent almost 2 hours thinking about an exercise that looked like it would take only 15 minutes and even tried pasting lemma (it's a good thing Professor Tu has solutions unlike ): The result of all that thinking is that I don't think $[0,1) \cup \{2\}$, $(\varepsilon,1) \cup \{2\}$ or $\{2\}$ is homeomorphic to any open subset of $\mathscr H^1$ or $\mathscr L^1$. I was able to show $\{0\} \subseteq \partial M$ and $(0,1)\subseteq M^0$, but I don't quite know where $2$ belongs. I believe $M$ is not locally $\mathscr H^1$. Also, I have double checked: I believe "manifold boundary" was defined for manifolds with boundary, so this isn't some trick where "manifold boundary" is actually defined for a Hausdorff and second countable space that need not be locally $\mathscr H^n$.     To generalize,     Is a half-open interval and a point not in the interval's closure a manifold with boundary?
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Does an open subset of $R^n$ exist that has dimension less than $n$ in the standard topology? All the less than $n$ dimensional subsets I can think of are not open.
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Most textbooks introduce differentiable manifolds via . This has the advantage of being concrete, but the disadvantage that the local coordinates are usually completely irrelevant- the choice of atlas and chart is arbitrary, and rarely if ever seems to play any role in differential geometry/topology.  There is a much better definition of differentiable manifolds, which I don't know a good textbook reference for, via . This definition does not involve any strange arbitrary choices, and is coordinate free. Paragraph 3 in Wikipedia (which is the actual definition) states:   A differentiable manifold (of class $C_k$) consists of a pair $(M, \mathcal{O}_M)$ where $M$ is a topological space, and $\mathcal{O}_M$ is a sheaf of local $R$-algebras defined on $M$, such that the locally ringed space $(M,\mathcal{O}_M)$ is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$.   This confuses me, because I don't see why such a sheaf should be acyclic, or where conditions like "paracompact" or "complete metric space" or "second countable Hausdorff" are implicit. So either:   The wikipedia entry has a mistake (I would want to sanity-check this before editing the entry, because this is such a fundamental definition which thousands must have read). Somewhere in that definition, the condition that $M$ be paracompact is implicit.   Question: Should the definition above indeed require that $M$ be second-countable Hausdorff or paracompact or whatever? Or is it somehow implicit somewhere, and if so, where?    Also, is this definition given carefully in any textbook?  Update: I have editted the Wikipedia article to require that $M$ be second-countable Hausdorff. But I'm still wondering if there is a textbook covering this stuff, and whether requiring the sheaf to be acyclic might have worked instead as an alternative.
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These are two definitions in page 48 of the book an introduction to manifolds by Loring Tu.  Definition 5.1. A topological space $M$ is locally Euclidean of dimension $n$ if every point $p$ in $M$ has a neighborhood $U$ such that there is a homeomorphism $\phi$ from $U$ onto an open subset of $\mathbb R^n$.  Definition 5.2. A topological manifold is a Hausdorff, second countable, locally Euclidean space. It is said to be of dimension $n$ if it is locally Euclidean of dimension $n$.  In the last lines of page 48, we reed,  Of course, if a topological manifold has several connected components, it is possible for each component to have a different dimension.  But this is a bit strange for me. If a topological manifold has several connected components and each component has different dimension, then how this manifold can be locally Euclidean space, say for example of dimension $n$? That is, by the above definition of topological manifolad, can a non-connected toplogical space be a topological manifold?
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Suppose $ T \subset \mathbb{C} $. Show that the corresponding set $ S \subset \Sigma $ is  a. a circle if $ T $ is a circle. b. a circle minus (0, 0, 1) if $ T $ is a line.  Here we are defining $ \Sigma $ to be the Riemann sphere, given by the set: $$ \Sigma = \left \{(\xi, \eta, \zeta) : \xi^{2} + \eta^{2} + (\zeta - \frac{1}{2})^{2} = \frac{1}{4} \right \} $$  To take a point from $ \mathbb{C} $ to $ \Sigma $ we can use the following:  $$ \xi = \frac{x}{x^{2} + y^{2} + 1}; \eta = \frac{y}{x^{2} + y^{2} + 1}; \zeta = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1} $$  We define a circle on $ \Sigma $ to be the intersection of a plane of the form $ A\xi + B\eta + C\zeta = D $ with $ \Sigma $. We also know the converse of this problem is true, that the intersection above yeilds a set in $ \mathbb{C} $ with the following property:  $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. As you can see, when C = D, then an equation for a line is yeilded, otherwise it is a circle.   I really am at a  loss about how to solve this problem. The only thing I can think to do is to pick 3 points on a circle or radius $ r $ with center $ z_{0} $, use these points to find two vectors in $ \Sigma $, take their cross product to get a normal vector, use this normal vector to get a plane. Once I have the plane in form $ A\xi + B\eta + C\zeta = D $ then I could prove that the circle I had chosen corresponds exactly with $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. Is there not an easier, less computation way to do this?
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I was wondering if there is a reason why in the definition of a topologic space, the reunion of any collection of open sets is also an open set but the intersection has to be finite to be a set.  Is there because of the definition of an open set itself ? I have always seen it as a definition, but now I want to know why.  Thanks for your help.
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We know that in the usual topology, countable or arbitrary intersection of open sets can zoom into a singleton, hence is not in the topology.   I am curious if there is well known classes of topologies, whose open sets maintains their openness after countable or arbitrary intersection.   I know two examples:      1) In the discrete topology, $\mathcal{T} = \mathcal{P}(X)$. The power   set is a $\sigma$-algebra, hence "closed" under countable   intersections, so a countable intersection of open sets in the   discrete topology is open.      2) In the particular point topology, given $x_o \in \mathbb{R}$,   $\mathcal{T}_{x_o} = \{U \subseteq \mathbb{R}| x_o \in  U\}\cup\{\varnothing\}$, an arbitrary intersection of open sets will   always contain $x_o$, in particular $x_o$ is open. Hence a countable   intersection of open sets in the particular point topology topology is   open.   Are there a lot more well known classes of topology whose condition on intersections of open set can be extended?
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We have the classic result that an open subset of $\mathbb{R}$ is an at-most-countable union of disjoint open intervals.  In trying to prove this, I found the following line of reasoning, and I would like to know if its correct, because it's quite different from the other proofs I have read.    Let S be an open subset of R.  Thus for every $a$ in $S$, $a$ is contained in some open ball $B$ which is contained in $S$.  In particular $a$ is contained in an open interval which is contained in $S$.  For each $a$, let's associate the maximal such interval $I_a=(\alpha, \beta)$ where $\alpha$ and $\beta$ live in $\mathbb{R} \bigcup \{\pm\infty\}$  $\alpha=\inf\{y: (y,a] \subset S\}$  $\beta=\sup\{y: [a,y) \subset S\}$  Now we have that $S$ is the union of $I_a$ over all $a$ in S. Furthermore if $I_{a_1}$ and $I_{a_2}$ are distinct, then they must be disjoint (otherwise we would contradict the maximality of at least one of them).   Hence $S$ is a disjoint union of open intervals.   Finally, each such interval contains a rational number which is not contained in any other such interval; if there were uncountably many intervals in the union, there would be uncountably many rationals, a contradiction.  Thus $S$ is an at-most-countable union of disjoint open intervals.
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Let $X$ be infinite set and $a \in X $. Let $$\mathscr{F} = \{ A \subset X: |A| \; \; finite \; \;, a \notin A \}$$  Then $\mathscr{F}$ is the set all of all closed sets for some topology $\mathcal{T} $ on $X$. Also, any two continuous maps $f,g: (X, \mathscr{T}) \to Y$ ( $Y$ is metric space) have the property that if $f(b) = g(b)$ for all $b \neq a $, then $f(a) = g(a)$.  In my notes my professor posted this theorem without proof, so I was wondering what is best way to proof this theorem.
989
What is an example of an open map $(0,1) \to \mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map $e^{1/z}$ from $\mathbb{C}$ to $\mathbb{C}$ (filled in to be $0$ at $0$). There must be a simpler example, using the usual Euclidean topology, right?
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An open set in an $n$-manifold is clearly a submanifold of the same dimension as its containing manifold (see open manifolds).   Now, given an $n$-manifold $M$, is it true that a set, to be the underlying set of a submanifold of $M$ with dimension $n$, must be open?