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As the topic, how can I prove it?

Prove that (a,b) and [a,b) in R are not homeomorphic metric space.

I'm pretty weak on this section (homeomorphism) and I don't have any clue to figure it out.

Please help me, thank you.

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Without going into formalisms, if two spaces are homeomorphic then all of their topological properties are the same. Now the space $(a,b)$ has the (topological) property that if you remove any of its points, the resulting space is disconnected. But the space $[a,b)$ does not have this property -- why?

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    $\begingroup$ because I can remove the point "a" from [a,b) and the resulting space is still connected. $\endgroup$ – Vito Chou Dec 6 '14 at 17:25
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    $\begingroup$ I can understand this discussion, makes me more familiar to homs. But I can't tell the professor this way. I'll try to make this discussion into a complete proof. $\endgroup$ – Vito Chou Dec 6 '14 at 17:29
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    $\begingroup$ By my lights, this is a complete proof. $\endgroup$ – Lubin Dec 6 '14 at 19:21
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    $\begingroup$ @Lubin: Between mathematicians already familiar with topology, it’s a prefectly good high-level proof. But in the context of, say, a first course in topology, it should probably have a few more details filled in. $\endgroup$ – Peter LeFanu Lumsdaine Dec 6 '14 at 22:47
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Thank to all above, now I write it this way and it seems complete.

Suppose not, i.e. if there exist $f:(a,b) \to [a,b)$ a homeomorphism.

By definition, f is continuous, $f^{-1}$ is continuous, and $f$ is bijective.

$f$ is bijective then we have the inverse image of $[a,b)$ is $(a,b)$

$f^{-1}(a)$ is a single point, say c, and $a \lt c \lt b$.

Consider $f^{-1}((a,b))=(a,c) \cup (c,b)$ is disconnected.

It contradicts the fact that $f^{-1}((a,b))$ is connected, noted that $(a,b)$ connected and $f^{-1}$ is continuous.

Therefore, $(a,b)$ and $[a,b)$ are not homeomorphic.

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  • $\begingroup$ You could have defined your function from $[a,b)$ to $(a,b)$. This way, the proof would have looked cleaner. $\endgroup$ – Prince Kumar Dec 3 '16 at 12:59
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Let $f: [a,b) \to (a,b)$ be a homeomorphism. Then the image of $(a,b)$ is $(a,b) \setminus \{f(a)\}$. Because $(a,b)$ is connected and $f$ is continuous, the image of $(a,b)$ is connected.

Can $(a,b) \setminus \{f(a)\}$ be connected (we know that $f(a)$ is in $(a,b)$ )? Assume it is. Now $(a,b)\setminus \{f(a)\} = (a, f(a)) \cup (f(a) ,b)$. What does this mean?

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    $\begingroup$ (a,f(a))and(f(a),b)are disjoint nonempty open intervals (open interval at least contains an open ball) and hence separate (a,b)\{f(a)}. Then it's disconnected. $\endgroup$ – Vito Chou Dec 6 '14 at 17:48

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