Continuity of the modulus function on $\mathbb R$.

Question

PROVE OR DISPROVE:

Let $f : \mathbb R \to \mathbb R$ be defined by $f(x) = |x|$. Then f is Tu − Tu continuous, where Tu is usual topology.

So we have to show that if $V$ is an open set belong to Tu then $f^{-1}(V)$ belongs to Tu.

I had hard time showing that $v$ satisfies that.

For example, if I take $V=(0,1)$ then $f^{-1}(V)=[0,\infty)$, but this does not belong to Tu. Hence , the statement is false

Is that right?

I believe $f^{-1}(0,1) = (-1,0)\cup (0,1)$, which is in Tu. Could you please explain why you think $f^{-1}(0,1) = [0, \infty)$? — Kenny Wong, May 02 '17 at 01:45
I think because f(x)=|x| which is positive value include zero — rian asd, May 02 '17 at 01:59
When approaching a question like this, it's always a good idea to start by reviewing the definition! :) The notation $f^{-1}(V)$ means the "preimage of $V$", i.e. $f^{-1}(V) = { x \in \mathbb R : f(x) \in V }$. So $f^{-1}(0,1) = { x \in \mathbb R : |x| \in (0,1) } = (-1,0) \cup (0, 1)$. — Kenny Wong, May 02 '17 at 02:09

score 1 · Answer 1 · answered May 02 '17 at 01:28

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Hint: Check what is the preimage of the open interval $(a,b)$ in $\mathbb{R}_{cod}$ (the set of reals in the codomain). Then realize that the set of all open intervals form a basis for the codomain.

answered May 02 '17 at 01:28

AspiringMathematician

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Continuity of the modulus function on $\mathbb R$.

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