$S=\left\{\dfrac{x^{2}}{1+x^{2}}:x \in \mathbb R\right\}$
Since $S$ is not closed (the limit point $1$ does not belong to the set), so I concluded that $S$ is not compact.
I am confused about verifying whether it is connected or disconnected.
S can also be written as $S=[0,1)$. (Is this correct?)
$[0,1)$ is a connected set, as it is not a subset of the union of any two disjoint nonempty open sets.
Thus, $S$ is NOT compact but connected.
I want to understand if my reasoning is correct, and if there is a better way to identify compact sets and/or connected sets. Thanks for your help..
Best Answer
From the fact that $S$ is not closed, you should have concluded that $S$ is not compact.
Connectedness (in fact, path-connectedness) can be verified without explicitly rewriting $S$ in a more comprehensible way (i.e., without recognizing that $S=[0,1)$. Note that $S$ is of the form $S=\{\,f(x)\mid x\in\mathbb R\,\}$ where $f$ is a continuous function. Then between any two points $s_0=f(t_0)$ and $s_1=f(t_1)$ of $S$ we always have a path $[0,1]\to S$, $\tau\mapsto f(t_0+\tau\cdot(t_1-t_0))$.
Of course, once you did identify $S$ as $[0,1)$ you should immediately recall (as you did in your reasoning) that such a half-open interval is not compact and that every interval is connected.