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It has been proved that $[0,1]$ is compact iff it is connected. I want to know

$\color{black}{\text{Can one deduce that $(0,1)$ is not compact then it is not connected?!!}}$

Of course the answer is certainly "No". But why? It is strange to me a bit!!

$\color{black}{\text{Do you have any mathematical theorem in your mind based on this logic?(just for comparing)}}$

Some related posts:

C.F.G
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3 Answers3

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$[0,1]$ is compact and connected so the equivalence $([0,1]$ is compact $\Longleftrightarrow [0,1]$ is connected) is true.

$(0,1)$ is not compact but connected so the implication $((0,1)$ is not compact $\Longrightarrow (0,1)$ is not connected) is false.

TheSilverDoe
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What is certainly untrue is that a topological space is connected if and only if it is compact. This is untrue even for subsets of the real line: consider $(0,1)$ and $[0,1]\cup[2,3]$, for instance.

If $X$ is some fixed topological space, then there isn't much content to the statement 'if $X$ is not compact then it is not connected'. Either $X$ is connected or it is not, and either $X$ is compact or it is not. As TheSilverDoe points out, $(0,1)$ is not compact but connected, so the implication is false in this case.

What the linked answer shows is that you can use compactness of $[0,1]$ to prove connectedness of $[0,1]$ and vice versa; however, this proof makes essential use of the whole structure of $[0,1]$, including the fact that it has 'endpoints'. It would be useful for you to go through the proof deriving compactness of $[0,1]$ from its connectedness and see why it fails for $(0,1)$.

John Gowers
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  • So can one say that $(0,1)$ is open iff it is connected? – C.F.G Sep 01 '20 at 09:25
  • Its model is "$p\iff q$" which implies "$\sim p\iff \sim q$". But I know that $\sim (0,1)\neq [0,1]$. – C.F.G Sep 01 '20 at 09:27
  • @C.F.G You can say that if you like. But 'if and only if' statements are only usually interesting if they are preceded by a quantifier: i.e., 'for all spaces $X$ of some particular form, $X$ is compact if and only if it is connected'. – John Gowers Sep 01 '20 at 10:04
  • What is the correct negative form of statement mentioned in title? (pls add this to your answer) – C.F.G Sep 01 '20 at 10:17
  • @C.F.G Sorry, I don't understand what you're asking for. – John Gowers Sep 01 '20 at 10:37
  • I meant what is the form $\sim p\iff \sim q$ for "[0,1] is compact ⟺ it is connected!"? – C.F.G Sep 01 '20 at 10:38
  • $[0,1]$ is not compact if and only if it is not connected? Is that what you mean? – John Gowers Sep 01 '20 at 10:40
  • Yes, if we simplify it, it become "[0,1] is not compact if and only if it is disconnected". But I thought that this is the correct one: "$[0,1]^c$ is not compact if and only if it is disconnected" – C.F.G Sep 01 '20 at 10:47
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    @C.F.G The main issue of your question is that you are using implication and equivalence between statements that we know are true or false. An implication can be useful if the two parts are sometimes true, sometimes false. Here, we can say logically that the equivalence is true, but it is very clumsy since the two parts are always true. – TheSilverDoe Sep 01 '20 at 18:04
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If $X$ has the order topology (from some linear order $<$ on $X$) then if $X$ is compact it has the lub-property (every subset $A \neq \emptyset$ that is bounded above $\sup(A) \in X$ exists. We cannot conclude however that $X$ is connected (as e.g. $\{0\} \cup \{\frac1n\mid n =1,2,3\} \subseteq \Bbb R$ has the order topology (in its inherited order), or also any finite subset of $\Bbb R$, and both are compact and disconnected). We also know that a compact ordered space has a minimum and a maximum (which is not needed for connectedness).

If however $X$ is a linearly ordered topological space, then $X$ connected implies $X^+$ is compact (where $X^+$ is $X$ with a minimum and/or maximum added when it didn't have it, so e.g. $(0,1)^+=[0,1] = [0,1)^+ = (0,1]^+$ etc. The connectedness gives almost enough of order-completeness by its lub property, only min or max could be missing.

So it's not iff within ordered topological spaces; more of them are compact than connected, and connected is "almost" compact. So the notions are linked that way.

Henno Brandsma
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