Proving that graph of $\sin(\frac{1}{x})$ is path connected in $\mathbb{R}^2$

Question

It seems visually clear to me that $A:=\{(x, \sin(\frac{1}{x})) : x\in (0, 1]\}\subset \mathbb{R}^2$ is path connected. But I'm struggling to make this precise.

Given $(a,b), (c,d) \in A$ with $a

Answer 1

All you need to know is that the image of a (path) connected space via continous function is again (path) connected.

Recall that for $f:X\to Y$ be continous the graph is defined as

$$Gr(f)=\{(x, f(x))\ |\ x\in X\}$$

Now we have a continous function

$$F:X\to X\times Y$$ $$F(x)=(x, f(x))$$

If $X$ is (path) connected then $\mbox{im}(F)=Gr(f)$ is (path) connected.

In your case your function is given by

$$f:(0, 1]\to\mathbb{R}$$ $$f(x)=\sin\bigg(\frac{1}{x}\bigg)$$

Answer 2

Hint

Consider $\varphi \colon [0,1] \to A$ $$ \varphi(t)= \left(a+(c-a)t, \; \sin \frac{1}{a+(c-a)t}\right).$$ Now argue that $\varphi$ is continuous.

Answer 3

Let $D\subset\mathbb{R}$ and $f_1,\ldots,f_n:D\to\mathbb{R}$ be continous functions. Let $f=(f_1,\ldots,f_n)$, i.e. $f:D\to\mathbb{R}^n,\ x\mapsto (f_1(x),\ldots,f_n(x))$.

Every open set in the product space $\mathbb{R}^n$ is the union of sets of the form $U_1\times\cdots\times U_n$ with $U_i$ open in $\mathbb{R}$. Also

$$f^{-1}(U_1\times\cdots\times U_n) = f_1^{-1}(U_1)\cap\cdots \cap f_n^{-1}(U_n)$$

is the intersection of finitely many open subsets in $\mathbb{R}$ thus open itself. This makes $f$ continous.

In your case $n=2$, $D=[a,c]$, $f_1:D\to\mathbb{R},\ t\mapsto t$ and $f_2 = \sin \circ g$ with $g:D\to\mathbb{R},\ t\mapsto \frac{1}{t}$. $f_1$ is continous and since $\sin$ and $g$ are continous, $f_2$ is, too.

Answer 4

A function $f: X \rightarrow Y \times Z$ is continuous iff $\pi_Y \circ f: X \rightarrow Y$ and $\pi_Z \circ f: X \rightarrow Z$ are both continuous, where $\pi_Y : Y \times Z \rightarrow Y ,\pi_Z: Y \times Z\rightarrow Z$ are the standard projections: $\pi_Y(y,z) = y, \pi_Z(y,z) = z$

For $f(x) = (x, \sin(\frac{1}{x})): [0,1) \rightarrow \mathbb{R}^2$, we have $\pi_1 \circ f =: [0,1) \rightarrow \mathbb{R}$ is given by $(\pi_1 \circ f)(x) = x$ which is surely continuous (as $[0,1)$ has the subspace topology w.r.t. $\mathbb{R}$) , and $\pi_2 \circ f(x) = \sin(\frac{1}{x})$ is also continuous on $[0,1)$, as the composition of the continuous $x \rightarrow \frac{1}{x}$ and $x \rightarrow \sin(x)$ (even differentiable).

So $f$ is continuous and your set is the image under $f$ of the path-continuous $[0,1)$, so also path-continuous.