 So, this is another way of proving the same thing. So, we have got as far as connectedness is concerned, we have analyzed the image, you can say image of a connected subset of real line is connected, because connected subsets are only intervals and the image is connected, so that is interval. So, another way of saying is continuity preserves connectedness, image of connected sets are, so let us write f d to r continuous, d connected, another way of writing implies f of d connected. Next, we want to analyze compactness, whether like connectedness, whether compactness is preserved or not. Let us look at a very special case, we know that compact subsets of real line are those which are closed and bounded, but let us look at a very special example when the compact set is a closed bounded interval. So, f is a function on a b to r continuous, we want to analyze what can we say about f of a b, the range of, you can also interpret it this way. We took intervals, image of an interval was an interval, now we have specialized the interval to be a closed bounded interval and so the image of this is going to be interval, because a b is an interval. The question is, is it a closed bounded interval or not? Earlier theorem said, f is continuous, i is interval, f of i is interval, now we are specializing, if it is a closed bounded interval, can I say f of i is a, so we know this, call it as some set a, we know a is an interval, so the question is, is a a closed bounded interval, is it a closed bounded interval? It is an interval we know. Now let us, a is closed, let us analyze one of them, a is closed, so what I have to show, if a is closed what is to be shown? Whenever I take a sequence in a, converging somewhere that image must be in a, so let a n belong to a, a n converge to a, to show a belongs to a. What does a n belong to a imply? Which is equal to f of a b implies a n is equal to f of x n for some x n belonging to a b. So, implies by compactness, x n has the sequence x n as a convergent subsequence, so let us write that also, x n k converging to a point, there is no x u somewhere, so x belonging to a b. So, we are transferring the problem from the domain range to the domain and doing property there now back, so I go back, implies f continuous f of x n k converging to f of x, f continuous, but where does f of x n k converge? f of x n is a n, a n is converging to a, so where does the subsequence converge? To a, same limit, so implies f of x is equal to a, limit is same, because x n k is a subsequence of x n, that means a n k corresponding a n k will be a subsequence of a n, a n converges to a, so that has to converge to the same limit. So, that means whenever I take any point in a, if a n converges to a, then a must be having a pre-image, that means a belongs to the range. So, implies a belongs to f of a b, that is equal to a, so a is close, a is bounded, I have already shown it is close, we have to show a is bounded, that means the range, if I take the function a b in the close bounded interval a b, the range has to be a interval we already know, it has to be a close interval we know, we only want to show it is bounded. If not, if the range of a function is not bounded means what? That means in the range there are points where you can go away and away from something. So, let us write a is bound, suppose not, then there exists, if not there exist points a n belonging to a b, say that f of a n, let us write mod f of a n goes to infinity. If the range is not bounded, there must be a sequence in the domain, which is either going to plus infinity or minus infinity. So, let us say mod of that goes to plus infinity, is that okay? Now a ns belong to a b, so again compactness, a n must have a convergent subsequence converging in a b. So, a n belonging to a b implies there it is a n k a subsequence a n k converging to a belonging to a, oh sorry. I should not write a because a is the end point. So, something else x, x belonging to a, but I have got continuity. So, f of x n must, f of a n k must converge, but can f of a n converge because the original sequence is okay? Or if you like divergent, let me write to be more precise, instead of this let us write is bigger than n for every n. Let us write that way, then it will make it very clear. If it is not bounded, I can find points where value is bigger than, otherwise it will be less than or equal to bounded. So, at a n the value is bigger than mod n. So, what is the value at f of a n k? It will be more precise than bigger than n k, which is bigger than k anyway. So, for every k absolute value of a n k, so it cannot converge, but other side it should converge to implies f of a n k cannot converge, but this star, star plus continuity implies f a n k is convergent. So, that is the contradiction, because a n k converges to x, f is continuous. So, f of that must converge, but here it is bigger than n k, bigger than k, so it cannot converge. So, that is the contradiction. So, we assume if not, that means if not bounded, then there is a problem. Suppose not, so that must be not true implies a is bounded. So, what we have said that if the domain of the function is a closed bounded interval, then the range also is a closed bounded interval. Can I say something more about that closed bounded interval? What that should be? Anybody can guess?