Let $f:[a,b] \to \mathbb{R}$ be a non-constant continuous function. Show that $f ([a,b])$ is uncountable.
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3Do you know the intermediate value theorem? – Ethan Bolker Apr 25 '17 at 17:20
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Obviously. Why not ? – gobinda chandra Apr 25 '17 at 17:27
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But how can i say f is uncountable? – gobinda chandra Apr 25 '17 at 17:29
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"$f$ is uncountable" makes no sense. You want to show that the set of values of $f$ is uncountable. – Ethan Bolker Apr 25 '17 at 17:30
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Yes yes . I want to mean f([a,b]) . – gobinda chandra Apr 25 '17 at 17:31
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But how can I do this ? – gobinda chandra Apr 25 '17 at 17:32
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1The proof is one or two sentences using the intermediate value theorem. If that's not enough of a hint you'll have to wait until someone decides to provide the answer for you. – Ethan Bolker Apr 25 '17 at 17:33
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Since f is non-constant without lose of any generality let a <b be two points so that f (a) <f (b) . Then IVT \thereexists c\in mathbb {R} s.t. f (a)<c <f (b). Hence there are uncountably many such c and f maps every such c. Thau f ([a,b]) is uncountable. Is it right ,Sir? – gobinda chandra Apr 25 '17 at 17:52
3 Answers
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One elegant approach is to use topological fact, that continuous image of connected set is connected. Connected subsets of $\mathbb{R}$ are precisely intervals and singletons. Since $f$ is not constant, $f([a,b])$ is nondegenerated interval, and thus uncountable.
More elementary approach is to use intermediate value theorem as mentioned by others. By assumption there exists $x_1,x_2\in [a, b]$ such that $f(x_1)<f(x_2)$. By intermediate value theorem if $y\in (f(x_1),f(x_2))$ then there exist $x\in(x_1,x_2)\cup(x_2,x_1)$ such that $f(x)=y$. Note, that $y$ was arbitrary, and $(x_1,x_2)\cup(x_2,x_1)\subset[a,b]$, so $[f(x_1),f(x_2)]\subset f([a,b])$. Therefore $f([a,b])$ is uncountable.
Przemek
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Hint: use Darboux Property. If a continuous function attains two values $y_1,y_2$, then ... ?
szw1710
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@egreg Are you sure? Maybe. I had in mind Intermediate Value Theorem. In Poland it is sometimes called Darboux Property. So, both continuous functions and derivatives do admit Darbous property. Please look at this: https://www.encyclopediaofmath.org/index.php/Darboux_property – szw1710 Apr 25 '17 at 17:37
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Hmm.. I seem to have the following argument. Suppose it is not uncountable. Let us denote $$ f([0,1]) \triangleq \{x_n:n \in \mathbb{N}\}. $$ Now, using the continuity of $f(\cdot)$, note that each of $A_n\triangleq f^{-1}(\{x_n\})$ is closed. But then, $$ [0,1] = \bigcup_{n=1}^{\infty}f^{-1}(\{x_n\}) = \bigcup_{n=1}^{\infty}A_n $$ would hold. After this point, note that $A_n$'s are all disjoint. So it boils down arguing whether $[0,1]$ can be written as a countable union of disjoint closed sets.
Apparently, the answer is no, and given in this link : Is [0,1] a countable disjoint union of closed sets?.
TBTD
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