# There exists a continuous function $f: [0,1] \rightarrow [0,1]$ $-$ {$\frac{1}{2}$} such that $f$ is on to. [closed]

There exists a continuous function $$f: [0,1] \rightarrow [0,1] - \left \{ \frac{1}{2} \right\}$$ such that $$f$$ is onto.

Is the above statement true?

I need some help to solve this. I am not sure how to work on it. Can anyone give me a detailed solution please! Thanks

• No. Does not exist. – Michael Rozenberg Sep 16 at 12:18
• Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. – 5xum Sep 16 at 12:18
• Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. Even if the question is closed, you can still edit it, and we will vote to reopen it. – 5xum Sep 16 at 12:18
• For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Martin R Sep 16 at 12:20
• And to edit your question, you can click on the word "edit" in light gray just below the question itself. (I mention this because it took me 6 months to notice that!) – John Hughes Sep 16 at 12:27

If $$f: [0,1] \to [0,1]$$ is continuous, then the image $$f([0,1])$$ is a compact interval, hence $$f([0,1]) \ne [0,1] \setminus \{1/2\}.$$