Let $X$ be a connected topological space such that there exists a non constant continuous function $f$:$X$$\to$$\Bbb{R}$ where $\Bbb{R}$ equipped with the usual topology. Let $f(X)$ $=$ {$f(x)$ | $x$ $\epsilon$ $X$ }. Then

$A$) $X$ is countable but $f($X$)$ is uncountable.

$B$) $f(X)$ is countable but $X$ is uncountable.

$C$) Both $f(X)$ and $X$ are countable .

$D$) Both $f(X)$ and $X$ are uncountable.

What I feel is since $X$ is connected and $f$ is continuous $f(X)$ must be connected.

Also $\Bbb{R}$ is equipped with usual topology and since $f$ is non constant continuous function from connected space;

$f(X)$ can not be countable ( since in $\Bbb{R}$ connected sets are intervals (uncountable) or they are just singleton (whic is not possible since $f$ is non-constant) ). So the options $(B)$ and $(C)$ are incorrect.

If we assume option $(A)$ is correct then we will get one-many map which will not be a function ( since cardinality of countable set (here $X$) will be less than cardinality of uncountable set ( here $f(X)$) ). hence I think option $(A)$ is also incorrect.

and hence option $(D)$ must be correct that both $X$ and $f(X)$ are uncountable.

please tell me whether my answer is correct or not. And also mention alternate method if any. Thanks.


notice that $f(X)$ must be a connected subspace of $\mathbb R$ that is not a point. (So it is an interval with more than one element). This implies $f(X)$ is uncountable.

In general, for any function $f$ we have $|X|\geq |f(X)|$. Therefore $X$ is also uncountable.

  • 1
    $\begingroup$ So yes, your argument is correct. Although I feel that you could have explained a bit more why a connected set that is not a point is uncountable. $\endgroup$ – Jorge Fernández Hidalgo Jan 23 '17 at 15:35
  • $\begingroup$ thank you so much for pointing out this point, I will try to edit question asap. :) $\endgroup$ – Math_Explorer Jan 23 '17 at 15:40

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