Equivalent definitions for $f:E \to \mathbb{R}$ being continuous?

Question

I would like to either prove or disprove the following. Essentially, I would like to see if two definitions for continuity (for a function $f:E \to \mathbb{R}$) are equivalent. The $ε$-$δ$ definition should be equivalent to Ross' definition for continuity in his Elementary Analysis (2nd Ed, p. 124). The second definition is motivated by Royden's definition for a continuous function from one topological space to another. (Real Analysis, 1st Ed, p. 126).

Let $f:E \to \mathbb{R}$ where $E\subseteq \mathbb{R}$.

Let $τ:=\{X\subseteq \mathbb{R} : X$ is an open set in $\mathbb{R}$ (under the usual metric) $\}$

Let $τ_E:= \{U \cap E : U \in τ\}$.

Then the following are equivalent:

• $\forall x_o \in E \forall ε>0 \exists δ>0 \forall x \in E[|x-x_0|<δ \to |f(x)-f(x_0)|<ε]$
• $\forall U\in τ[f^{-1}(U) \in τ_E]$

I tried proving the "forward direction" that the first definition implies the second as follows. Let $U \in τ$. Ιf $f^{-1}(U)$ is empty, we're done, so suppose not. Take $x_0 \in f^{-1}(U)$. Then $f(x_0) \in U$ and so there is a $ε>0$ such that $(f(x_0)-ε,f(x_0)+ε)\subseteq U$. Choose $δ>0$ such that $\forall x \in E[|x-x_0|<δ \to |f(x)-f(x_0)|<ε]$. Then $(x_0-δ, x_0+δ) \cap E \subseteq f^{-1}(U).$

In the special case that $E=\mathbb{R}$, it is easy to see that $x_0$ must be an interior point of $f^{-1}(U)$. Since $x_0$ was chosen arbitrarily, this would imply that $f^{-1}(U)$ is open in $T_ \mathbb{R}$ if $E=\mathbb{R}$

However, in the general case, I do not see how to prove that $f^{-1}(U) \in τ_E$.

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The "backward" direction, however, I have been able to prove as follows. Let $x_0 \in E$ and $ε>0$. We have $f^{-1}((f(x_0)-ε,f(x_0)+ε)) \in τ_E$. Write $f^{-1}((f(x_0)-ε,f(x_0)+ε))=O \cap E$ for some $O \in τ$.

$x_0 \in O \cap E$, so there is a $δ>0$ such that $(x_0-δ,x_0+δ) \subseteq O$.

So $\forall x \in E [x \in (x_0-δ,x_0+δ) \to x \in O \cap E]$.

So $\forall x \in E [|x-x_0|<δ \to f(x) \in (f(x_0)-ε,f(x_0)+ε)]$ and we're done.

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I have been able to prove the backward direction but am having trouble with the forward direction.

Answer 1

If we are able to show that $f^{-1}(U)=V\cap E$ where $V$ is an open set in $\mathbb{R}$, then we are done.

Clearly you were able to show that for every $x_{0}\in f^{-1}(U)$, $\exists\ \delta_{x_{0}}>0$ such that $$(x_{0}-\delta_{x_{0}},x_{0}+\delta_{x_{0}})\cap E\subseteq f^{-1}(U)$$ $$\therefore f^{-1}(U)\subseteq \cup_{x_{0}\in f^{-1}(U)}(x_{0}-\delta_{x_{0}},x_{0}+\delta_{x_{0}})\cap E$$ $$\therefore f^{-1}(U)=\cup_{x_{0}\in f^{-1}(U)}(x_{0}-\delta_{x_{0}},x_{0}+\delta_{x_{0}})\cap E$$ Let $V=\cup_{x_{0}\in f^{-1}(U)}(x_{0}-\delta_{x_{0}},x_{0}+\delta_{x_{0}})$ which is open in $\mathbb{R}$. $$f^{-1}(U)=V\cap E$$