$S^{1}$ is a subset of Euclidean plane $R^2$ equipped with the relative topology. I have a set:
$$G = \{(x,y) \in R^2 | x^2 + y^2 = 1 \ y>0\} \cup \{(1,0)\}$$
How to prove that G is not open or closed set of $S^{1}$?
The definition of induced topology states that if G is open set then $$G = U \cap S^{1}$$ where $U$ is an open set in $R^2$. This should produce a contradiction since G is not an open set. How to deduce a contradiction in this situation?
On
If $\epsilon >0$ is sufficiently small then $(\sqrt {1-{\epsilon}^{2}},-\epsilon)$ is in U. It is also on the circle so it must be in G. But this contradicts the definition of G. Hence G is not open. The proof of the fact that G is not closed is similar since the complement of G is similar to G.
Any open set in $\Bbb{R}^2$ that contains $(1, 0)$ will also contain neighboring (possibly arbitrarily close) points. But of course, these neighboring points would lie outside $G$, contrary to the definition of an open set.