# Real Analysis: Continuity & Intervals

• lmn123
In summary, we use the theorem from the book stating that a continuous function with a compact domain is bounded and has a greatest and least point in its image. We then apply the intermediate value theorem to show that any value between these two points is also in the image, making the image an interval. In the case of a half open interval, we can use a similar approach by finding the max and min values of subintervals within the half open interval. Overall, this shows that the image of a continuous function with an interval domain is also an interval.

## Homework Statement

If the domain of a continuous function is an interval, show that the image is an interval.

## Homework Equations

Theorem from book: f is a cont. function with compact domain D, then f is bounded and there exists points y and z such that f(y) = sup{ f(x) | x ∈ D} and f(z) = inf{ f(x) | x ∈ D}
The intermediate value theorem.

## The Attempt at a Solution

Case 1: D=[a,b]
Since D is compact and by hypothesis, f is continuous we know that there exist points y and z in D such that f(y) = sup{ f(x) | x ∈ D} and f(z) = inf{ f(x) | x ∈ D}. By definition of sup and inf these are the greatest and least points of the image of f. And so, by the intermediate value theorem, we know that any t in between f(z) < t < f(y), there exists an x in D such that f(x)=t. So, we can write the interval of the image as [f(z),f(y)].

Case 2: D=(a,b)
We can write (a,b) as an infinite union of closed sets. For example ∪[a+1/m, b-1/m] from m=1 to infinity.
By the previous case we know that each [a+1/m, b-1/m] attains a sup and inf value for all 1/m. Now we can simply take the max( sup{ f(x) | x ∈ [a+1, b-1]}, sup{ f(x) | x ∈ [a+(1/2), b-(1/2)]}, sup{ f(x) | x ∈ [a+(1/3), b-(1/3)}, ... ) and call it p. Similarly we can take the min( inf{ f(x) | x ∈ [a+1, b-1]}, inf{ f(x) | x ∈ [a+(1/2), b-(1/2)]}, inf{ f(x) | x ∈ [a+(1/3), b-(1/3)}, ... ) and call it q. This will give us an open interval in which all values of the image lie. And so, the interval of the image of f is (q,p).

Is this the right way to prove this? It seems right to me but I also have my doubts haha. Thanks!

Kreizhn said: