Prove that $f$ is continuous on $H$ if and only if... Let $f :H \rightarrow \mathbb{R}$ be a function. Prove that $f$ is continuous on $H$ if and only if $f(\overline{T}) \subseteq \overline{f(T)}$ for every subset $T$ of $H$.
This is what I was thinking...
Let $f :H \rightarrow \mathbb{R}$ be continuous and let $T$ be a subset of $H$. Then, clearly $f(T) \subset \overline{f(T)}$. Therefore, $T \subset f^{-1}(f(t)) \subset f^{-1}(\overline{f(T)})$. Since $\overline{f(T)}$ is a closed subset of $\mathbb{R}$ then $f^{-1}(\overline{f(T)})$ is closed in $H$. Thus, $f$ is continuous. Now $T \subset f^{-1}(\overline{f(T)}) \Rightarrow \overline{T} \subset f^{-1}(\overline{f(T)}) \Rightarrow f(\overline{T)} \subset \overline{f(T)}$ for every subset $T$ of $H$. 
Now consider $f(\overline{T}) \subset \overline{f(T)}$ for every subset $T$ of $H$. Let $x$ be a closed subset of $\mathbb{R}$. Then $\overline{x} = x$. Now, $f^{-1}(x)$ is a subset of $H$. Then, $f(\overline{f^{-1}(x)}) \subset \overline{f(f^{-1}(x)} \Rightarrow f(\overline{f^-1(x)}) \subset  \overline{x} = x$. So, $\overline{f^{-1}(x)} \subset f^{-1}(x) \Rightarrow f^{-1}(x) = \overline{f^{-1}(x)}$. Therefore, $f^{-1}(x) \subset \overline{f^{-1}(x)}$ and thus $f^{-1}(x)$ is closed in $H$. So $f$ is continuous. 
This this correct?
 A: For the first direction, I think you can show it in an easier way. 
Take, $x$ which is a limit point of $T$, and let the sequence in $T$, $x_n\to x$. By continuity of $f$, $f(x_n)\to f(x)$. Hence, $f(x)\in \bar {f(T)}$.
A: There are several equivalent definitions of continuity, for a general metric/toplogical space. I assume $H$ here is an arbitrary metric/topological space and not the upper half plane sometimes denoted by $\mathbb{H}$. 
If $H$ is a metric space, the answer by the user $\textbf{Juanito}$ as elucidated above works just fine. 
If $H$ is an arbitrary toplogical space without any metric specified. Then one has to make sense of "$\rightarrow$", i.e the notion of limit. 
But we could use the standard definition of continuity, i.e $f^{-1}(C)$ is closed for every closed subset $C \subseteq \mathbb{C}$. 
"$\Rightarrow$" : $f$ is continuous then the above property holds. 
Let $T$ be an arbitrary set, and $\overline{T}$ be it's closure (intersection of all closed subsets of $H$ containing $T$) and $\overline{f(T)}$ be the closure of $f(T)$. 
Since $f$ is continuous $f^{-1}(\overline{f(T)})$ is a closed subset of $H$ which contains $T$, therefore must contain $\overline{T}$. i.e $f(\overline{T}) \subseteq \overline{f(T)}$ as required.
"$\Leftarrow$" : Conversely, if $f(\overline{T}) \subseteq \overline{f(T)}$ $\forall \ T \subseteq H$ then $f$ is continuous. 
Let $C$ be a closed subset of $\mathbb{C}$. 
Since, $f(\overline{T}) \subseteq \overline{f(T)}$ holds $\forall \ T \subseteq H$, it must also hold for $f^{-1}(C)$
$\implies$ $f(\overline{f^{-1}(C)}) \subseteq \overline{f(f^{-1}(C))}\subseteq C$  (because $f(f^{-1}(C)) \subseteq C$)
Therefore $\overline{f(f^{-1}(C))} \subseteq C$, since $C$ is closed by assumption.
$\implies$ $\overline{f^{-1}(C)}\subseteq f^{-1}(C) \subseteq\overline{f^{-1}(C)}$, implying $f^{-1}(C) = \overline{f^{-1}(C)}$. 
We have proved that inverse image of a closed set is closed which is an equivalent definition of continuity valid for arbitrary topological spaces.
