The inverse image of an open set is open

[jfy4]

Homework Statement

Consider a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ and an arbitrary open $U\subset\mathbb{R}$. Show that the inverse image under $f$ of $U$, [itex]f^{-1}[/itex], is open.

Homework Equations

The definitions of open sets and continuity

The Attempt at a Solution

Pick an arbitrary point in the set [itex]f^{-1}[/itex], [itex]x_0\in f^{-1}[/itex]. Then $f(x_0)\in U$ which implies that $f(x_0)$ is an interior point. Then there exists a $\epsilon>0$ such that $V_{\epsilon}(f(x_0))$ exists. Then since the function is continuous, there exists a $\delta>0$ such that for all $x\in(x_0-\delta,x_0+\delta)\implies f(x)\in V_{\epsilon}(f(x_0))$. Then $(x_0-\delta,x_0+\delta)\subseteq U$ which means there exists a [itex]V_{\delta}(x_0)\subseteq f^{-1}[/itex]. Then $x_0$ is an interior point, but $x_0$ was arbitrary so every point in [itex]f^{-1}[/itex] is an interior point. Hence, [itex]f^{-1}[/itex] is open. $\blacksquare$

 

[Stephen Tashi]

What does the notation $V_{\epsilon}(f(x_0))$ mean? An open interval of width $$\epsilon$$ that contains $$f(x_0)$$ and is a subset of $$U$$?

 

[jfy4]

What does the notation $V_{\epsilon}(f(x_0))$ mean? An open interval of width $$\epsilon$$ that contains $$f(x_0)$$ and is a subset of $$U$$?

It means a neighborhood around $f(x_0)$ of width $2\epsilon$. The interval $(f(x_0)-\epsilon,f(x_0)+\epsilon)$

 

[jfy4]

It should say [itex]V_{\delta}(x_0)\subseteq f^{-1}[/itex] for starters, not a subset of U

EDIT: I fixed it.

 

[Stephen Tashi]

You need to say that $$V_\epsilon(f(x_0))$$ can be chosen to be a subset of $$U$$. You aren't making it clear why the continuity of $$f(x)$$ makes it possible to find $$V_\delta(x_0)$$. Does your text define continuity as a property involving a limit? If so, you need to explain why the definition allows you to find $$V_\delta$$.

 

[jfy4]

You need to say that $$V_\epsilon(f(x_0))$$ can be chosen to be a subset of $$U$$. You aren't making it clear why the continuity of $$f(x)$$ makes it possible to find $$V_\delta(x_0)$$. Does your text define continuity as a property involving a limit? If so, you need to explain why the definition allows you to find $$V_\delta$$.

Okay. $V_\epsilon$ is a subset since by definition an interior point has a neighborhood around it that is completely contained in the mother set.

I am given that the function is continuous. Then

$$(\forall\epsilon>0)(\exists\delta>0)(\forall x)(|x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon )$$.

I invoke the definition since I have found a neighborhood around $f(x_0)$ so I'm guaranteed a $\delta$-neighborhood where all its members map into $U$. Then the result follows.

If I were to include something like this in the proof would that be acceptable?

 

[Stephen Tashi]

Okay. $V_\epsilon$ is a subset

You mean that it may be chosen to be a subset of $$U$$.

"since by definition an interior point has a neighborhood around it that is completely contained in the mother set.

Yes, that's relevant after you have explained why $$f(x0)$$ is an interior point of $$U$$.

I am given that the function is continuous. Then

$$(\forall\epsilon>0)(\exists\delta>0)(\forall x)(|x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon$$.

OK, if that's how your text defines continuity. But the more common definitions are:

$$f(x)$$ is continuous at the point $$x_0$$ means $$lim_{x\rightarrow x_0} f(x) = f(x_0)$$
and
$$f(x)$$ is continuous on the set $$S$$ means that for each point $$x_0 \in S$$, $$f(x)$$ is continuous at $$x_0$$.

In those definitions there is nothing said about epsilons or deltas. The epsilons and deltas come from applying the definition of limit to the definition of continuity.

(An amusing fact:
In courses on topology, the definition "f is continuous" is that for the mapping f, the inverse image of each open set is open. This is because in General Topology, no measurement of distance is used to define "open" set, so there are no epsilons and deltas available to define limits or continuity.)