Is this space Pathconnected? Is the infinite unions of  $\cup_{n=1}^{\infty} (\frac{1}{n},1)  \cup\{{0}\}$ path connected.
Are they actually equal to the space $[0,1)$?
Thanks
 A: Yes, it is path-connected and equal to $[0,1)$.
A: Yes, this is just $[0,1)$, as
$$ \bigcup_{n=1}^{\infty} (\frac{1}{n},1) = (0,1)$$
A: Certainly it is true that $\bigcup_{n=1}^{\infty} (\frac{1}{n},1)  \cup\{{0}\} \subseteq [0,1)$.
What about the other direction? Pick any $x \in (0,1)$ (we already know that $0$ is contained in the left-hand side). By the Archimedean Property, there is $N$ such that $0 < \frac{1}{N} < x$. Thus, $x \in (\frac{1}{N}, 1)$. Since $x$ was arbitrary, $\bigcup_{n=1}^{\infty} (\frac{1}{n},1)  \cup\{{0}\} \supseteq [0,1)$, and so the two sets are equal.
A: to prove that $\cup_{n=1}^{\infty} (\frac{1}{n},1)  \cup\{{0}\}=[0,1)$ we just prove the double inclusions:
If $X\in\cup_{n=1}^{\infty} (\frac{1}{n},1)$ then there exists $n\in\mathbb{N}^*$ such that $X\in (\frac{1}{n},1)$ then $X\in [0,1)$ because $0<\frac{1}{n}\leq 1$ if $X=0$ then $X\in [0,1)$.
If $X\in [0,1)$ and $X\neq 0$ then by the property of infinimum, there exists $n\in\mathbb{N}^*$ such that $X\in (\frac{1}{n},1)$ then $X\in \cup_{n=1}^{\infty} (\frac{1}{n},1)$ and if $X=0$ then $X\in \cup_{n=1}^{\infty} (\frac{1}{n},1)  \cup\{{0}\}$.
we conclude that $\cup_{n=1}^{\infty} (\frac{1}{n},1)  \cup\{{0}\}=[0,1)$, and for sure $[0,1)$ is path connected. 
