Arbitrarily close uniformly continuous functions

Question

While preparing for my Real Analysis exam I tried to solve the following problem I found:

Problem:

Let $A \subseteq \mathbb{R}$ and suppose that $f: A \rightarrow \mathbb{R}$ has the following property: for each $\varepsilon>0$ there exists a function $g_{\varepsilon}: A \rightarrow \mathbb{R}$ such that $g_{\varepsilon}$ is uniformly continuous on $A$ and $\left|f(x)-g_{\varepsilon}(x)\right|<\varepsilon$ for all $x \in A .$ Prove that $f$ is uniformly continuous on $A$.

Attempt:

Given $A \subset \mathbb{R}$, $f: A \rightarrow \mathbb{R}$ and $\epsilon > 0$ one can obtain a function $g_{\varepsilon}$ which is uniformly continuous on $A$ such that: $$ |f(x)-g_{\varepsilon}(x)|<\frac{\epsilon}{3}, \forall x \in A $$

Given $y \in A$, one can also obtain that: $$ |f(y)-g_{\varepsilon}(y)|<\frac{\epsilon}{3}, \forall y \in A $$

Additionally, since $g_{\varepsilon}$ is uniformly continuous, $\forall \epsilon^{'} > 0$ $\exists \delta > 0$ such that $\forall x, y \in A$ $$ \left|g_{\varepsilon}(x)-g_{\varepsilon}(y)\right|<\varepsilon / 3 $$ whenever $|x - y| < \delta$

Now, considering that $$|f(x) - f(y)| = $$ $$= |f(x) - g_{\varepsilon}(x) + g_{\varepsilon}(x) - g_{\varepsilon}(y) + g_{\varepsilon}(y) - f(y)| \leq $$

$$\leq |f(x) - g_{\varepsilon}(x)| + |g_{\varepsilon}(x) - g_{\varepsilon}(y)| + |g_{\varepsilon}(y) - f(y)| < \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}$$ $$= \epsilon$$

whenever $|x - y| < \delta$

Therefore, we conclude that f is uniformly continuous.

Questions:

1. Is the solution correct?

2. If the solution is indeed correct, is it well written?

3. If the solution is incorrect, can someone gently me explain why and provide a solution?

Thanks in advance, Lucas!

Answer 1

As Kavi Rama Murthy said, your proof is good. The write up is also generally okay, but does have some issues. These don't affect the quality of the proof, but make it harder to read. (One trivial thing you can do is decide which epsilon you prefer and stick with it. Mixing $\epsilon$ and $\varepsilon$ as you do is a little distracting.)

Given $A \subset \mathbb{R}$, $f: A \rightarrow \mathbb{R}$ and $\varepsilon > 0$ one can obtain a function $g_{\varepsilon}$ which is uniformly continuous on $A$ such that: $$|f(x)-g_{\varepsilon}(x)|<\frac{\varepsilon}{3}, \forall x \in A$$

Good.

Given $y \in A$, one can also obtain that: $$|f(y)-g_{\varepsilon}(y)|<\frac{\varepsilon}{3}, \forall y \in A$$

There are two problems here. First you introduce an actual variable with the phrase "Given $y \in A$". But then you make no further use of it. No, really. It is not used anywhere in the proof. All of those other "$y$" variables are quantified (that "$\forall y \in A$" is called a relative quantifier - drop the "$\in A$" and it would be a universal quantifier). By quantifying those other uses of $y$, you've made them dummy variables. They only have meaning inside the range of each of their quantifications. Outside those ranges, there is no $y$. Other than this one unnecessary introduction.

The second problem is why Kavi Rama Murthy told you the second sentence should be deleted: This $$|f(x)-g_{\varepsilon}(x)|<\frac{\varepsilon}{3}, \forall x \in A$$ and this $$|f(y)-g_{\varepsilon}(y)|<\frac{\varepsilon}{3}, \forall y \in A$$ are exactly the same statement. The $x$ and $y$ in them are dummy variables, which are variables used only in support of a notation. The meaning of the notation did not change in the slightest because you used a different letter. So all you are doing in this line is repeating what you said immediately above it.

Additionally, since $g_{\varepsilon}$ is uniformly continuous, $\forall \varepsilon' > 0\ \exists \delta > 0$ such that $\forall x, y \in A$ $$\left|g_{\varepsilon}(x)-g_{\varepsilon}(y)\right|<\varepsilon / 3$$ whenever $|x - y| < \delta$

Other than the unused "$\forall \varepsilon'$", which can be deleted, this is good.

Now, considering that $$|f(x) - f(y)| = $$ $$= |f(x) - g_{\varepsilon}(x) + g_{\varepsilon}(x) - g_{\varepsilon}(y) + g_{\varepsilon}(y) - f(y)| \leq $$ $$\leq |f(x) - g_{\varepsilon}(x)| + |g_{\varepsilon}(x) - g_{\varepsilon}(y)| + |g_{\varepsilon}(y) - f(y)| < \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}$$ $$= \varepsilon$$

There are no logic issues here. That both $$|f(x) - g_{\varepsilon}(x)| < \frac{\varepsilon}{3}$$ and $$|f(y) - g_{\varepsilon}(y)| <\frac{\varepsilon}{3}$$ follow from the single earlier statement of $|f(x)-g_{\varepsilon}(x)|<\frac{\varepsilon}{3}, \forall x \in A$ since both the $x$ value used here (which is a different variable than the dummy variable in the statement) and the $y$ value used here are in $A$, they satisfy the $\forall$ condition.