Check if the given set is Connected and Compact.

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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.

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Deepabali Roy
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Deepabali Roy

Updated on August 01, 2022

Comments

  • Deepabali Roy
    Deepabali Roy over 1 year

    $S=\left\{\dfrac{x^{2}}{1+x^{2}}:x \in \mathbb R\right\}$

    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..

  • Deepabali Roy
    Deepabali Roy almost 9 years
    Oops that compact thing was a typing error.
  • Deepabali Roy
    Deepabali Roy almost 9 years
    @Hagen Does path-connectedness always imply connectedness? Also could you please explain how you defined the path?