Images of intervals

We can define an interval to be a convex subset I of R. That is, it has the property that if a, b ∈ I and a ≤ c ≤ b then c ∈ I.

Theorem

The image of an interval under a continuous map is also an interval.

Proof

If f (a), f (b) ∈ f (I) and y lies between them, then by the intermediate value theorem there is an x between a and b with f (x) = y.

Recall that intervals come in various types.

Open: like (a, b), (a, ∞), (-∞, ∞), ...

Closed: like [a, b]

Half-open: like [a, b), (a, b], [a, ∞), ...

Theorem

The image of a closed, bounded interval under a continuous map is closed and bounded.

Proof

By the theorem of the previous section, the image of an interval I = [a, b] is bounded and is a subset of [m, M] (say) where m, M are the lub and glb of the image. Since the function attains its bounds, m, M ∈ f (I) and so the image is [m, M].

Remark

The images of any of the other intervals can be any kind of interval.

1. Any open interval can be mapped to any other open interval (in fact by a continuous bijection):

   For example, (-π/₂, π/₂)→ (-∞, ∞) by the tangent function.

   (-∞, ∞)→ (0, ∞) by the exponential function.

   Any finite open interval can be mapped to any other finite open interval by a suitable linear function. (Exercise)

2. Any half-open interval can be mapped to any other half-open interval using the same maps as in 1.

3. Any open or half-open interval can be mapped to a closed interval.

   For example, (-∞, ∞) or [0, ∞)→ [-1, 1] by the sine function.

   Any open interval can be mapped to a half-open interval.
   For example, (-∞, ∞)→ [0, ∞) by the x² function.
   

4. Finally, (the tricky one) any half open interval can be mapped onto an open interval.
   For example, [0, ∞)→ (-∞, ∞) by the map x→ x sin x.