• No products in the cart.

# Image of continuous function (TIFR 2015 problem 2)

Question:

Let $$f: \mathbb{R} \to \mathbb{R}$$ be a continuous function. Which of the following can not be the image of $$(0,1]$$ under $$f$$?

A. {0}

B. $$(0,1)$$

C. $$[0,1)$$

D. $$[0,1]$$

Discussion:

If f is the constant function constantly mapping to 0, which is continuous, then the image set is {0}.

Suppose that $$f((0,1])=(0,1)$$ . Then $$f((0,1))=(0,1)- \{f(1)\}$$. Now since $$f(1)\in (0,1)$$ the set $$(0,1)- \{f(1)\}$$ is not connected. But $$(0,1)$$ is connected, and we know that continuus image of a connected set is connected. This gives a contradiction. So $$(0,1)$$ can not be the image of $$(0,1]$$ under f.

Define $$f(x)=1-x$$.  Then $$f((0,1])= [0,1)$$.

Define $$f(x)=0$$ for $$x\in [0,\frac{1}{2}]$$ and $$f(x)= 2(x-\frac{1}{2})$$ for $$x\in [\frac{1}{2} ,1 ]$$. $$f$$ is continuous on  $$(0,\frac{1}{2}]$$ and $$[\frac{1}{2} ,1 ]$$ and $$f$$ agrees on the common points, by pasting lemma $$f$$ is continuous on $$[0,1]$$ . And image of $$(0,1]$$ is $$[0,1]$$.

November 27, 2017