Critique my proof of $f(x) = x^2, x \in \mathbb{R}$ being continuous

Question

Let $\epsilon > 0$ and $a \in \mathbb{R}$. Let $\delta > 0$ s.t $\delta < \frac{\epsilon}{\vert x + a \vert}$ and $\lvert x - a \rvert < \delta$. Thus,  $$\lvert f(x) - f(a) \rvert = \lvert x^2 - a^2 \rvert = \lvert x + a \rvert \lvert x - a \rvert < \delta \lvert x + a \rvert$$$$ < \frac{\epsilon}{\lvert x + a\rvert} \lvert x + a \rvert = \epsilon.$$Hence, $\lvert x - a \rvert < \delta \Rightarrow \lvert f(x) - f(a) \rvert < \epsilon$. Because $a$ was arbitrary, we conclude that $f(x) = x^2, x \in \mathbb{R}$ is continuous. 

Answer 1

The error lies in the sentence

Let $\delta > 0$ s.t $\delta < \frac{\epsilon}{\vert x + a \vert}$ and $\lvert x - a \rvert < \delta$.

As I commented, the choice of $\delta$ cannot depend on $x$ because there is no variable $x$ in scope at this point.

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Let's review the strategy for crafting an $\epsilon$-$\delta$ continuity proof. Given the function $$ f(x) = x^2, \quad x \in \mathbb{R}, $$ our goal is to show that

$f$ is continuous,

which is by definition equivalent to

for any $a \in \mathbb{R}$, $f$ is continuous at $a$.

Fix $a \in \mathbb{R}$. By the definition of continuity, we need to show that

for any $\epsilon > 0$, there exists a number $\delta > 0$ such that, for any $x$, $$ \lvert x - a \rvert < \delta \quad \implies \quad \lvert x^2 - a^2 \rvert < \epsilon. $$

Fix $\epsilon > 0$. To show that there exists a number $\delta > 0$ such that the aforementioned condition holds, we pick a value for $\delta$ based on quantities that are in scope (namely $a$ and $\epsilon$) and show that the desired condition holds for that value of $\delta$.

To do so, we observe that the desired relation $\lvert x^2 - a^2 \rvert < \epsilon$ is equivalent to $\lvert x - a \rvert \lvert x + a \rvert < \epsilon$, so we would like to bound both factors $\lvert x - a \rvert$ and $\lvert x + a \rvert$ from above. Automatically, the first factor $\lvert x - a \rvert$ is bound by whatever value of $\delta$ we choose. There is some freedom in how we choose to bound the second factor $\lvert x + a \rvert$; one possibility is to take advantage of the triangle inequality to get $$ \lvert x + a \rvert = \lvert (x - a) + 2a \rvert \le \lvert x - a \rvert + \lvert 2a \rvert < \delta + 2 \lvert a \rvert. $$ If we choose $\delta$ such that, for example, $$ \delta \le 1, \tag{1} \label{1} $$ then $$ \lvert x + a \rvert < 1 + 2 \lvert a \rvert, $$ so that in order to guarantee $\lvert x - a \rvert \lvert x + a \rvert < \epsilon$, it suffices to make $$ \lvert x - a \rvert < \frac{\epsilon}{1 + 2 \lvert a \rvert}. \tag{2} \label{2} $$ Therefore, we decide to pick $$ \delta = \min \Bigl\{ 1, \frac{\epsilon}{1 + 2 \lvert a \rvert} \Bigr\} $$ in order to guarantee both \eqref{1} and \eqref{2}.

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This is, of course, not the only possible choice of $\delta$. For example, Thomas Andrews's answer mentions an alternative choice $$ \delta = \sqrt{\epsilon + a^2} - \lvert a \rvert. $$

Answer 2

It pays well, to think of the definition of continuity in terms of neighbourhoods.

Definition. Let $f:A \to \mathbf{R}$ be any real-valued function. The function $f(x)$ is said to be continuous at $c \in A$, if for all $\epsilon > 0$, there exists $\delta > 0$, such that for all $x \in A$, satisfying $|x - c| < \delta$, the distance $|f(x) - f(c)| < \epsilon$.

Mathematically, $f(x)$ is continuous at $c$, iff

$$(\forall \epsilon > 0)(\exists \delta)(\forall x \in A)(\forall |x-c|<\delta)(|f(x) - f(c)|<\epsilon)$$

Equivalently,

Definition. The function $f(x)$ is said to be continuous at $c \in A$, if no matter what $\epsilon$-neighborhood of $f(c)$ you choose, there exists a corresponding $\delta$-neighboorhood around $c$, such that if ($x$ is in the domain $A$ and) $x$ lies in $V_\delta(c)$, you find that $f(x)$ lies in $V_\epsilon(f(c))$.

Mathematically, $f(x)$ is said to be continuous at $c$ iff

$$(\forall V_\epsilon(f(c))(\exists V_\delta(c))(\forall x \in A)(\forall x\in V_\delta(c))(f(x)\in V_\epsilon(f(c)))$$

You could have a different $\delta$-response to each $\epsilon-$ challenge and the point $c$. But, this $\delta$-response holds for all $x$ falling in $V_\delta(c)$.

Answer 3

You are correct to assume we want a useful bound on $|x+a|$ when $|x-a|<\delta.$

But you cannot choose $\delta$ in terms of $x.$

But you can use: $|x+a|\leq |x-a|+2|a|.$

Then: $$|x^2-a^2|\leq |x-a|^2+2|a||x-a|.$$

So you want: $$\delta^2+2|a|\delta\leq \epsilon.$$

Or: $$(\delta+|a|)^2\leq \epsilon+a^2$$ or $$\delta\leq \sqrt{\epsilon +a^2}-|a|$$

You have to prove the right side is positive, but that isn't hard.

Answer 4

It is always good practice to clearly state upfront all that you know from a problem:

$f:\mathbb R\to\mathbb R,\quad x\mapsto x^2$

Let $\:(x,a)\in\mathbb R\times\mathbb R\:$.

From the definition of a continuous function,

$$\forall x\:\:\forall \varepsilon > 0\:\: \exists \delta_{x,\varepsilon} > 0\:\text{ s.t. } |x - a| < \delta_{x,\varepsilon}\Rightarrow |f(x) - f(a)| < \varepsilon$$

So from the definition the idea is that $\:\delta\:$ could depend on both $\:x\:$ and $\:\varepsilon$.

Compare this to the uniform continuity definition:

$$\:\:\forall \varepsilon > 0\:\: \exists \delta_{\varepsilon} > 0\:\text{ s.t. }\:\forall x,y\:\:\: |x - y| < \delta_{\varepsilon}\Rightarrow |f(x) - f(a)| < \varepsilon$$