Prove that $\overline S$ is not path connected, where $S=\{x\times \sin(\frac1x):x\in(0,1]\}$

Question

The goal is to show that $\overline S$ is not path connected, where $S=\{x\times \sin(\frac{1}{x}):x\in(0,1]\}$.

Proof. Suppose there is path $f:[a,c]\to\overline S$ beginning at the origin and ending at a point of $S$. The set of those $t$ for which $f(t)\in0\times[-1,1]$ is closed, so it has a largest element $b$. Then $f:[b,c]\to\overline S$ is a path that maps $b$ into the vertical interval $0\times [-1,1]$ and maps the other points of $[b,c]$ to points of $S$. Replace $[b,c]$ by $[0,1]$ for convenience; let $f(t)=(x(t),y(t)).$

Then $x(0)=0,$ while $x(t)>0$ and $y(t)=\sin(1/x(t))$ for $t>0$.

We show there is a sequence of points $t_n\to0$ such that $y(t_n)=(-1)^n.$ Then the sequence $y(t_n)$ diverges, contradicting the continuity of $f.$

To find $t_n$ we proceed as follows: Given $n,$ choose $u$ with $0<u<x(1/n)$ such that $\sin(1/u)=(-1)^n$. Then use the intermediate value theorem to find $t_n$ with $0<t_n<1/n$ such that $x(t_n)=u.$

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• I don't understand how this function $f(t)=(x(t),y(t))$ is defined. $f(0)=(x(0),y(0))=(0,?)$

• About the sequence of points $t_n$, I don't understand the process of construction.

Given $n\in\mathbb N,$ let $u\in(0,x/n)\dots$ Is $x$ is a function right?

Using IVT, why the interval for $t_n$ is $(0,1/n)$?

Answer 1

This answer comes in three short sections that provide the details allowing one to unravel the 'mysteries' behind Munkres proof.

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Starting alongside of Munkres, we have a continuous function

$\tag 1 f:[a,c] \to \mathbb R \times \mathbb R \text{ such that } f(a) = (0,0) \text{ and } f(c) = s_0 \text{ with } s_0 \in S$

Munkres argues that there exists a unique number $b$ such that $a \lt b \lt c$ and $f(b) \in \{0\} \times [-1, 1]$ but $f$ applied to the interval $(b,c]$ lies inside of $S$. Let the $y\text{-coordinate}$ of $f(b)$ be denoted by $b_2$.

So we can restrict $f$ to $[b, c]$ and $f(b) = (0, b_2)$ and $f(c) = s_0$. A reparameterization can be found, so finally we have (using $f$ again):

$\tag 2 f:[0,1] \to \mathbb R \times \mathbb R \text{ such that } f(0) = (0,b_2) \text{ & } f(1) = s_0 \text{ with } s_0 \in S \text{ & } f(0,1] \subset S$

Now Munkres notes that the function $f$ in (2) can be written as $f(t) = (x(t), y(t))$.

Question: Is there a second reparameterization where $x(t) = t$? If yes, does this mean that we have a real valued continuous function

$\quad g: [0, +\infty) \to \mathbb R \text{ such that } g(x) = \sin(\frac{1}{x}) \text{ for } x \gt 0?$

The Munkres analysis can be compared with facts known about such functions $g$, defined by specifying the value $g$ takes at $x = 0$. See

$\quad$ Show $f(x)=\sin\frac1x$ is discontinuous on $\Bbb R$ using open balls

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Any function $f: [0,1] \to \mathbb R \times \mathbb R$ can be written in the form $f(t) = (x(t), y(t))$. If $f$ is continuous then both $\pi_x \circ f$ and $\pi_y \circ f$ are continuous, where $\pi_x$ and $\pi_y$ are the coordinate projection maps.

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The following is offered as a 'baby step' in understanding Munkres proof and might provide an 'intuitive feel' for the problem:

Now $S=\{(x\times \sin(1/x)):x\in(0,1]\}$ and for any $v \gt 0$

$\tag p S \; \; \bigcap \; \; (0, v] \times \{+1\} \;\ne \emptyset$

and

$\tag n S \; \; \bigcap \; \; (0, v] \times \{-1\} \; \ne \emptyset.$

To see this, note that the function $y = \sin(x)$ is periodic and 'keeps hitting' both $+1$ and $-1$.

The OP in encouraged to draw some pictures here that includes the shape of $f(t)$ described by (2) in Section 1.

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If $\sin(1/u)=(-1)^n \text{ then } 1/u = \arcsin((-1)^n). $

So

$u = (-1)^n (\frac{\pi}{2} + 2 \pi k)^{-1}.$

Answer 2

$f(0) = (x(0),y(0)) = (0,?)$

It doesn't matter. Take the value of $y(0)$ to be anything you like. (From the paragraph above, $f$ simply "maps $b$ into the vertical interval $0\times [-1,1]$".)

$t_n$

The point is essentially that we want to construct a sequence of points $t_n$ converging to $0$ (so the author has chosen $0 < t_n < 1/n$ for each $n$, which guarantees that $t_n\to 0$) but such that $y(t_n)$ doesn't converge at all. The $x$ here is indeed a function - it's the same function $x(t)$, just now with $t = 1/n$.

Answer 3

Here is a proof in a more topological style: For brevity let $T=\bar S \setminus S.$

Suppose $f:[0,1]\to \bar S$ is continuous with $f(0)\in T.$ For any $x\in [0,1]$ such that $f(x)\in T$ let $f(x)=(0,v)$ and let $$V(f(x))=\bar S\cap [(-1/2,1/2)\times (-1/2+v,1/2+v)].$$ Then $V(f(x))$ is a nbhd of $f(x)$ in the space $\bar S.$ By continuity of $f$ there exists $\delta >0$ such that $f$ maps $[0,1]\cap (-\delta+x,\delta +x)$ into $V(f(x)).$ For brevity let $$[0,1]\cap (-\delta+x,\delta+x)=U(x).$$ An inspection of the graph of $S$ shows that $V(f(x))\cap S =\cup W$ where $W$ is a countably infinite family of pair-wise-disjoint sets of the form $\{(d,\sin 1/d):d'<d<d''\},$ with $0<d'$, such that $\bar w_1\cap \bar w_2=\emptyset$ for any distinct $w_1,w_2 \in W.$ (Closure bar denoting closure in $\Bbb R^2$.)

Each member of $W$ is open-and-closed in the sub-space $V(f(x)).$ So, since the continuous image of the connected space $U(x)$ is connected, there is at most one $w_0\in W$ such that $\emptyset \ne w_0\cap f(U(x)).$

But if $w_0\in W$ and $\emptyset \ne w_0\cap f(U(x))$ then the two sets $w_0\cap f(U(x))$ and $T\cap f(U(x))$ are disjoint and not empty.... ( Note: $T\cap f(U(x))$ is not empty as it contains $f(x)\;$)....and are both open in the connected sub-space $f(U(x)),$ and their union is $f(U(x)).$ This contradicts the connectness of $f(U(x)).$ Therefore no such $w_0$ exists.

So $f(U(x))\cap w$ for all $w\in C.$ So $f(U(x))\subset T.$

We now have a kind of Heine-Borel argument: $f(0)\in T$ and if $f(x)\in T$ then $x$ has a nbhd $U(x)$ in the space $[0,1]$ such that $f((U(x))\subset T,$ from which we readily prove that $f([0,1])\subset T=\bar S \setminus S.$

So there cannot be a path from any member of $T$ to any member of $S.$

Answer 4

Let $S=\{(x, \sin(1/x)) \;:\; x\in(0,1]\}$ and $T = \{0\} \times [-1, 1]$. Then

$\tag 1 \overline S \text{ is equal to the disjoint union of } S \text{ with } T$

Lemma: Let $a \lt u \lt b$ and $f: (a, b) \to \overline S$ be a continuous function such that $f(u) \in S$.
Then the range of $f$ is included in $S$.

Proof (Sketch):

To arrive at contradiction assume that $f$ hits points in $T$. Now the set $S$ is open in $\overline S$ and its inverse image $U$ is therefore open and contains $u$. Moreover, the complement of $U$ is a nonempty closed subset of $(a,b)$. By employing logical symmetry, we assume now that we have a number $c \gt u$ such that $f(c) \in T$ and the image of the open interval $(u, c)$ under $f$ is contained in $S$. Since the left limit $\lim_{x \to c-}f(x)$ exists one can show that $f$ can't be going 'crazy' inside of $S$ as $x$ approaches $c$ from the left. More precisely, it can be shown that there exist an $n \ge 1$ such $f$ maps $(u,c)$ into the open set

$\tag 2 S_n =\{(x, ~\sin(1/x)) \;:\; x\in(\frac{1}{n},1]\}$

of $\overline S$, since otherwise we can argue that $f(c) = (0,+1)$ and $f(c) = (0,-1)$. But then of course $f(c)$ must be in $S$, a contradiction. $\quad \blacksquare$

Answer 5

Let $H=\{(x, \sin(1/x)) \;:\; x\in(0,1)\}$ and $T = \{0\} \times [-1, 1]$. Let

$\tag 1 S = H \cup T$

It is easy to show that $S$ is path-connected if and only if $S \cup \{\left(1,~\sin(1)\right)\}$ is path-connected. This is only mentioned because we modified (ever so slightly) the OP's question to fit the machinery found below.

The space $S$ is not path-connected.

Assume, to get a contradiction that we have a path $\gamma$ in $S$ connecting the point $(0,0)$ to $\left(\frac{1}{2}, ~\sin(\frac{1}{2})\right)$. The inverse image of $T$ under $\gamma$ is a closed subset of $[0,1]$ and has a maximum value not equal to $1$. By modifying this path, we can define another path

$\tag 2 \omega: [0,1] \to S$

satisfying $\omega(0) \in T$, $\,\omega(1) = \left(\frac{1}{2}, ~\sin(\frac{1}{2})\right)$ and $\omega\left( (0,1] \right) \cap T = \emptyset$.

But this path can also be regarded as connecting the two points inside the space

$\tag 3 G = H \cup \{ (0, \pi_y(\omega(0)) \}$ with $\pi_y$ the projection onto the $y\text{-axis}$

The concluding remark found here means that this is impossible.