A connected space that admits a nonconstant continuous map into reals is uncountable Let $f:X\to Y$ be a non-constant continuous map of topological spaces. 
If $Y=\mathbb{R}$ and $X$ is connected then $X$ is uncountable. True or False?

I know that $f(X)$ is an interval. The difficulty is in going from $X$  countable to $f(X)$   countable. 
 A: As $X$ is connected and $f$ is continuous so $f(X)$ is a connected subset of $\mathbb R$. So, $f(X)$ may be singleton set or any interval. But if $f(X)$ is a singleton set then the map $f$ is a constant map...contradiction. So, $f(X)$ must be an interval of $\mathbb R$. As , any interval is uncountable so, $f(X)$ is uncountable , which implies $X$ is uncountable.

Note : Let , $f:X\to Y$ be any map. Then , $X$ is countable $\implies$ $f(X)$ is countable.
Let , X be countable. As, f:X→f(X) is onto so if f(X) is uncountable then there is a point in f(X) which has no pre-image , fails to be onto.


A: We know that $f(X)$ is connected as a subspace of $\mathbb R$ since images of connected spaces under continuous maps are connected. Furthermore, there are at least two points $a, b \in f(X)$, such that $a \neq b$, because $f$ is non-constant. Assume without loss of generality that $a < b$. If $f(X)$ were countable, then there is some real number $c \notin f(X)$ such that $a < c < b$. Otherwise, then entire interval $(a, b)$, which is uncountable, would be contained in $f(X)$. Try to use this $c$ to obtain a separation of $f(X)$, contradicting the assumption that $X$ is connected. So $f(X)$ must be uncountable, which implies $X$ is uncountable (there are uncountably many elements in $f(X)$, and since each element in $f(X)$ is the image of some $x \in X$, there must be uncountably many $x \in X$).
