# A continuous surjective function from $(0,1]$ onto $$I’m trying to construct a continuous surjection from (0,1] onto [0,1], but I’m not getting anywhere. I don’t immediately see a contradiction which falsifies the existence of such a function, so my intuition tells me one exists. I feel like an absolute value function would work, but I’m not sure how to arrive at it in the proper way. Thanks for any help. #### Solutions Collecting From Web of "A continuous surjective function from (0,1] onto$$"

Try $f(t) = \sin(1/t)$.

If you want to use your absolute value idea, try making a $V$ shape with the vertex at $(\tfrac12, 0)$, opening upwards to include the points $(0,1)$ and $(1,1)$. (I’ll leave the actual function up to you.)

$$4\left(x-\frac{1}{2}\right)^2$$

Some point in the interior of $(0,1]$ will have to map to $0$. I picked $1/2$. Then the rest of the function has to never pass below $0$, so I squared it. Then it needs to actually cover up to and including $1$, so multiply by $4$. The collection of choices here should indicate that there are many such functions.

You could even cover this interval with only the domain $(0,1)$ using a similar method — minimum at 1/3, maximum at 2/3, cubic…

How about $f(x) = 2 |x-\frac{1}{2}|$

The function $f(x) =\sin^2 2\pi x$ or the variant with cosine should do. It is more than continuous: it is differentiable and has a power series representation.