quantchem said:I did this :
let
|f(x) - f(y) | < epsilon and |x - y| < delta. then epsilon < delta and therefore for each epsilon > 0, there is some delta > 0 such that |f(x) - f(y) | < epsilon when |x - y| < delta. so f is continuous at y when x is in an interval where f is defined.
Is this sufficient explanation?
Continuity in real analysis refers to the property of a function where small changes in the input result in small changes in the output. In other words, a function is continuous if its graph has no breaks or holes.
To prove continuity of a function in real analysis, one must show that the limit of the function as the input approaches a certain value is equal to the output at that value. This can be done using the epsilon-delta definition or the sequential criterion for continuity.
Pointwise continuity refers to the continuity of a function at each individual point in its domain. Uniform continuity, on the other hand, requires that the function remains continuous over an entire interval, rather than just at individual points.
The Intermediate Value Theorem states that if a function is continuous on a closed interval, then it takes on every value between the function's values at the endpoints of the interval. This theorem relies on the property of continuity to ensure that there are no breaks or gaps in the function's graph.
Yes, a function can be continuous at only one point in its domain. This is known as a removable discontinuity, where the function is defined at the point but has a hole or break in its graph. However, a function must be continuous at every other point in its domain to be considered continuous overall.