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a. This set is clearly disconnected because Determinant map is a continuous function from $\mathcal{O}_n(\mathbb{R})$ to the discrete set $\{1, -1\}$

c. This set can be thought as $\mathbb{R^2}-\mathbb{Q^2}$ which is connected as poved here Proving $\mathbb R^2 \setminus \mathbb Q^2$ is connected

b. I'm confused with this option and can't think of any possible ways to justify. Please give some hint. Thank you.

Dots_and_Arrows
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1 Answers1

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Hints for (b):

Let $f,g \in S$.

Let $p:[0,1]\to C[0,1]$ be defined by $$p(t)=(1-t)f+tg$$

Show that $p$ is continuous.

Show that the image of $p$ is a subset of $S$.

Note that $p(0)=f$, and $p(1)=g$.

Deduce that $S$ is path-connected.

quasi
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