How to prove that this function:

$$\begin{array}{ccc} f:[0,2] & \longrightarrow &[0,2] \\ x& \longmapsto & f(x) \\ \end{array},\; f(x)=\left\{\begin{array}{cc} 0& x\in [0,1], \\ x-1 & x\in [1,2]. \end{array} \right.$$

is continuous using the following definition? $$\forall V\in \mathcal{V}_{f(x)},\exists W\in \mathcal{V}_x, f(W)\subset V$$

Edit: I think that we must prove that $f$ is continuous in $x=1$

because it is clear that in $[0,1]$ and $[1,2]$ f is continuous. the problem is in $x=1$, and I don't know how to apply the definition to $x=1$.

Edit2: please why f is closed and not open ?

  • $\begingroup$ This is basically the epsilon delta definition for neighborhoods. What have you done so far? $\endgroup$
    – Alp Uzman
    Nov 29, 2015 at 9:00
  • $\begingroup$ @A.AlpUzman i edited the message i don't know how to apply the definition to x=1 $\endgroup$
    – Vrouvrou
    Nov 29, 2015 at 11:06

1 Answer 1


Let $\varepsilon > 0$, so that $[0,\varepsilon) = V \in \mathcal{V}_{f(1)}$. It suffices to show the condition of $V$ of this form.

Note that $f(1) = 0$, so there is no problem to the left of $x$ as there $f$ only assumes the value $0$. So take $W = (0, 1 + \varepsilon) \in \mathcal{V}_1$. Then if $x \le 0$ we know that $f(x) = 0 \in V$ and also if $x > 1$, then $f(x) = x - 1 < \varepsilon$, so $f(x) \in V$ as well, so $f[W] \subset V$ as required.

  • $\begingroup$ A ngbh of 1 must be $]1-\varepsilon,1+\varepsilon[$ no? i don't understand $\endgroup$
    – Vrouvrou
    Nov 29, 2015 at 12:00
  • $\begingroup$ Yes, but here even the (generally larger) neighbourhood starting from $0$ will work. You can also take $(1-\varepsilon, 1 + \varepsilon)$ if you like. $\endgroup$ Nov 29, 2015 at 12:28
  • $\begingroup$ please how to see that f is closed and not open ? $\endgroup$
    – Vrouvrou
    Dec 1, 2015 at 9:24
  • $\begingroup$ @Vrouvrou closed is automatically the case, as a continuous function from a compact space to a Hausdorff space. Not open because the image of $(0,1)$ is not open in $[0,2]$. $\endgroup$ Dec 1, 2015 at 13:22
  • $\begingroup$ We must work on the topology of [0,2] ? can we say that for example $f(]\frac12,\frac32[)=[0,\frac12[$ is not open on $(\mathbb{R},|.|)$ ? because i don't know how it is exactly the forme of an open on the topology of [0,2] , i know that is $I\cap[0,2]$ where $I$ is open in $\mathbb{R}$ that's all $\endgroup$
    – Vrouvrou
    Dec 1, 2015 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.