Topologist's sine curve is connected

I just came across the example of the topologist's sine curve that is connected but not path-connected. The rigorous proof of the non-path-connectedness can be found here.

But how can I prove that the curve is connected? To be honest, even intuitively I am not being able to see that the curve is connected. I am thinking if it is proved that the limit point of $\sin(1/x)$ as $x \to 0=0$, then it would be proved. But, why is this true? IMO, this limit doesn't exist. Intuitively also, it seems that the graph would behave crazily and not approach a particular value as a tends to $0.$

EDIT (Brett Frankel): There are a few different working definitions of the topologist's since curve. For the sake of clarity/consistency, I have copied below the definition used in the linked post: $$ y(x) = \begin{cases} \sin\left(\frac{1}{x}\right) & \mbox{if $0\lt x \lt 1$,}\\\ 0 & \mbox{if $x=0$,}\end{cases}$$

Solution 1:

If the graph $X$ of the topologist's sine curve were not connected, then there would be disjoint non-empty open sets $A,B$ covering $X$. Let's assume a point $(x,\ \sin(1/x))\in B$ for some $x>0$. Then the whole graph for positive $x$ is contained in $B$, only leaving the point $(0,0)$ for the set $A$. But any open set about $(0,0)$ would contain $(1/n\pi,\ \sin(n\pi))$ for large enough $n\in\mathbb N$, thus $A$ would intersect $B$.

Solution 2:

Call the topologist's sine curve $T$, and let $A = \{(x,\sin 1/x)\in\mathbb{R}^2\mid x\in\mathbb{R}^+\}$, $B = \{(x,\sin 1/x)\in\mathbb{R}^2\mid x\in\mathbb{R}^-\}$. Then $T \subseteq \overline{A\cup B} = \overline{A}\cup\overline{B}$. It isn't difficult to show that $A$ and $B$ are connected (even path connected!) and then you just need two lemmas to show that $T$ is connected:

Lemma 1: If $A\subseteq X$ is a connected subset of a metric space $X$ and $A\subseteq B\subseteq \overline{A}$, then $B$ is also connected. Edit: Stefan H. reminds us that this result also holds when $X$ is a general topological space, not just a metric space.

Lemma 2: If $A$ and $B$ are connected, and $A\cap B\neq\emptyset$, then $A\cup B$ is also connected.

And neither of these should be too hard to show (Hint: use the fact that if $X$ is connected and $f : X\to\{0,1\}$ is continuous, then $f$ is constant).