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Hint: Pick $x\in (0,1)$. To prove that $X$ is open in $\Bbb R$, you must show that there is a ball around $x$ that is completely contained in $(0,1)$. Let $\delta = \min(x,1-x)$, then $\delta$ is the smaller of the distances between $0$ and $x$, and $x$ and $1$. What happens if you consider $B(x,\delta)$? In the complex case, balls are disks. If you look at ...

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If an open set contains a limit point of $E$, then it must also contain a point of $E$ itself (by the definition of limit point), so showing it doesn't intersect $E$ is enough.

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By definition (of compactness), given any compact space $X$, any open cover of $X$ has a finite subcover. In general, however, one cannot say before one has an open cover how many open sets one might need to make a subcover. For example, for any $\epsilon > 0$, the set of $\epsilon$-balls in $[0, 1]$ is an open cover, and so admits some finite subcover. ...

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Using the converse you mentioned, it is clear that if $f$ or $f^{-1}$ fails to map a connected set to another connected set, $f$ cannot be a homeomorphism. Thus, it suffices to show that Theorem: Fix $n > 1$. Let $f:\mathbb{R}^n \to \mathbb{R}^n$ be a bijection, such that $f$ maps connected sets to connect sets, and $f^{-1}$ maps connected sets to ...

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Ideia: Take the sequence $x_n=(\frac{1}{n},1)$, it converges to $x_0$. If existed such homotopy $H(x,t)$ then the sequences $H(x_{n},t)$ would still converge to $x_{0}$. You have for each neighborhood $U$ of $x_0$ a number $N_{(U,t)}$ and $\epsilon_{(U,t)}$ such that $H(x_n,s)\in U$ for $n>N_{(U,t)}$ and $|t-s|<\epsilon_{(U,t)}$. Covering the ...

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Suppose $a \in A$ such that $\overline a \cap U \ne \varnothing$. Either $U$ contains a point of $a$ and we are done, or $U$ contains a limit point of $a$, but then $U$ is a neighborhood of a limit point of $a$, so it must contain a point of $a$ itself.

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