Why formulate continuity in terms of pre-images instead of image? I wanted to discuss my intuition of why we formulate the concept of continuity in terms of pre-image of open set is open instead of images for example if we consider $f(x) = c$ where $c$ is some constant, then that should be continuous, but if we formulated continuity in terms of images it won't be continuous. Is the reason we do it that way, because pre-images will always guarantee that function in the domain can only have $1$ value?
It's not nice to formulate it in terms of images. The behavior of images varies wildly between various continuous functions. Take the interval $(-10,10)$ and the functions $f_1(x) = x$ and $f_2(x) = \sin x$. Then the image of $(-10,10)$ under $f_1$ is of course $(-10,10)$, but the image of $(-10,10)$ under $f_2$ is $[-1,1]$. With one function you get an open set, with another you get a closed set. You can of course explore the other cases of half-open/half-closed or other phenomena. You can't make any meaningful statement about images of open sets. You can't even say anything for certain about local behavior (the images of "small" open sets) because there are plentiful locally constant functions.

You might want to flip this on its head then since images of open sets doesn't work: what about images of closed sets? This also doesn't work. Consider $\mathbb{R}$ and the function $f_3(x) = \arctan x$, then the image of $\mathbb{R}$ under $f_3$ is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ which is an open set.

The beauty of pre-images is that the functions we intuitively look at as being continuous have open sets as pre-images of open sets. Perhaps the closest you can get to a statement about images is that continuous functions map compact sets to compact sets, however there are discontinuous functions which also do this: consider $[-1,1]$ and $f_4(x) = 1$ if $x&gt;0$, $f_4(0) = 0$ and $f_4(x) = -1$ if $x&lt;0$. Then $f_4$ maps a compact set to a compact set but it is definitely not continuous in the usual topology.
I realize this is a bit of thread necromancy, but I don't see the harm, and I'm not sure I see this perspective noted in the answers yet.  It's true that defining it in terms of pre-images works, and defining it in terms of images (without changing anything else) doesn't work.  But why doesn't it work?
This puzzled me too, when I first encountered it mumblety-mumble years ago.  The way that I ended up internalizing it is that it's related to the fact that a function has a unique output for each input, but it doesn't have to have a unique input for each output.  Two different inputs can map to the same output.  In some sense, therefore, the output of an input set—its image, in other words—&quot;loses information&quot; about the input set, and therefore (in a non-rigorous sense), the image can't be expected to preserve the topology of its input.
That isn't the case with the pre-image.  It may have multiple &quot;copies,&quot; so to speak, of the information in the image—as for example with periodic functions—but each &quot;copy&quot; preserves the topology and corresponds, open set to open set, even for non-periodic functions.
Consider what happens when a continuous function $f:\mathbb{R} \to \mathbb{R}$ does take on the same output for multiple inputs.  So long as $f$ is (strictly) monotonic and continuous, then it will map an open set to an open set; the topology is preserved.  But if it isn't monotonic—if there's a local maximum, say, at $(x_0, f(x_0))$—then the image of some open neighborhood of $x_0$ cannot itself be open, because $f$ has to attain a maximum value of $f(x_0)$ in that neighborhood; it can't attain a maximum of &quot;the largest real value smaller than $f(x_0)$.&quot;
Contrast this with the situation of the pre-image.  Because you can't have a single input value map to multiple output values, there's no way for an open set in the co-domain to correspond to a non-open set in the domain; you'd have to have the input &quot;turn back on itself,&quot; which it can't do (so long as $f$ is a function).