Showing continuity by the inverse image of every open set is open for addition and product function.

Question

Here are the 2 questions:

I am wondering how can I Show continuity by using the following definition of continuity: the inverse image of every open set is open. could anyone help me in this please?

Answer 1

The Question tells you to draw the pre-image, I think it's quite obvious if you follow the instruction.

Q1:

It's a plane in $\mathbb{R}^3$. (See the following figure)

So no matter how you choose $(a,b)$, for each $m\in(a,b)$ the preimage of it will be the line $y=-x+m$ in $\mathbb{R}^2$. Thus the preimage of the whole interval is the area between $y=-x+a$ and $y=-x+b$ (i.e. doesn't contain the two boundary lines). Now, consider a sequence $(x_n)\in \sigma^{-1}((a,b)^C)$ converging to $x\in \sigma^{-1}((a,b)^C)$. Suppose not, then since $x_n\to x$, we can find $\epsilon>0$ s.t. $d(x_n,x)<\epsilon$ which means this sequence is in $\sigma^{-1}((a,b))$ contradicts our assumption, so $x\in \sigma^{-1}((a,b)^C))$ therefore $\sigma^{-1}((a,b))^C$ is closed in $\mathbb{R}^2\implies\sigma^{-1}((a,b))$ is open.

Q2: Similarly, follow the instruction, the preimage $\mu^{-1}((a,b))$ lies between two hyperbolas $xy=a$ and $xy=b$, using the same method, you can prove its continuity.

Answer 2

If $(x_0,y_0)\in\sigma^{-1}(J)$ or equivalently $a<x_0+y_0<b$ then you can find $\epsilon>0$ such that $a<x+y<b$ whenever $|x-x_0|<\epsilon$ and $|y-y_0|<\epsilon$.

This means that the open set $(x_0-\epsilon,x_0+\epsilon)\times(y_0-\epsilon,y_0+\epsilon)\subseteq\mathbb R^2$ contains $(x_0,y_0)$ and is a subset of $\sigma^{-1}(J)$.

Here $(x_0,y_0)$ is an arbitrary element of $\sigma^{-1}(J)$ so actually it has been shown that for every $(x,y)\in\sigma^{-1}(J)$ we can find a set $U_{(x,y)}$ open in $\mathbb R^2$ and satisfying $(x,y)\in U_{(x,y)}\subseteq\sigma^{-1}(J)$.

Then we can conclude that: $$\bigcup_{(x,y)\in\sigma^{-1}(J)}U_{(x,y)}=\sigma^{-1}(J)$$where the LHS is a union of open sets.

This justifies the conclusion that $\sigma^{-1}(J)$ is open.

This can be proved for every open interval $J$ so we are allowed to conclude that $\sigma$ is continuous.

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The same principle can be applied in order to prove that $\mu$ is continuous.