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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

4 Answers 4

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Yes, it is path-connected and equal to $[0,1)$.

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    @rk101 I'm sorry : )2012-03-26
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Yes, this is just $[0,1)$, as

$ \bigcup_{n=1}^{\infty} (\frac{1}{n},1) = (0,1)$

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    @rk101: That's a totally new question. Ask it separately if you want to ask it.2012-03-26
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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.

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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.