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Test your Math skills

March 17, 2007 – 7:04 pm

Here’s a question that was asked in a recent oral exam for a Master’s degree in Mathematics.

Let $f : [0,1] \rightarrow \mathbb R$ be a real function such that has a limit at each point. Does have at least one continuity point?

Everything you need to solve this question is covered in the first year or so of undergraduate math. Don’t continue reading if you want to try solving it yourself…

Solution. We will show that has a continuity point. has a limit at each point, so let $F : [0,1] \rightarrow \mathbb R$ be a function such that

$\forall x \in [0,1]$ $F(x) = lim_{y \rightarrow x} f(y)$

Lemma.

is continuous.

Proof. Choose some $x_0 \in [0,1]$ . Intuitively, F(x_0)

’s limit, so in a small enough surrounding of x_0

will be close to F(x_0)

. Hence, the limits F(x)

will also be close to F(x_0)

, and thus

is continuous at x_0

.

Formally, let $\epsilon > 0$ . Then there exists $\delta > 0$ such that

$\forall x \in (x_0-\delta,x_0+\delta) \setminus \{x_0\} \; |f(x) - F(x_0)|<\epsilon$

Which means that for all such x

$|lim_{y \rightarrow x} f(y) - F(x_0)|<\epsilon$

$|F(x) - F(x_0)|<\epsilon$
And the Lemma is proved.

Our purpose now is to show that there is a point

such that

. Let’s count the points of

that are ‘far away’ from the limit

. Choose some $\epsilon>0$ and define the set:

$A_\epsilon = \{ \, x \, | \, f(x) \, > \, F(x) + \epsilon \, \}$

Lemma. $A_{\epsilon}$ is finite.

Proof. Suppose $A_{\epsilon}$ is infinite, then because [0,1]

is compact we can find a series $\{x_n\} \subset A_\epsilon$ such that $x_n \longrightarrow x_0$ and $x_n \neq x_0$ for some $x_0 \in [0,1]$ .

Therefore, $f(x_n) \longrightarrow F(x_0)$ . But for large enough , we must have

$f(x_n) \, > \, F(x_n) + \epsilon \, > \, F(x_0) + {\epsilon \over 2}$
where we have used

’s continuity at x_0

. Thus we reach a contradiction, and $A_\epsilon$ must be finite.

Now let’s take $A = \bigcup_{n=1}^{\infty} A_{1/n}$ . Then from the lemma we have that A is an enumerable set. A also contains all the points for whom $f(x) \, > \, F(x)$ .

Likewise we can define

$B_\epsilon = \{ \, x \, | \, f(x) \, < \, F(x) - \epsilon \, \}$
$B = \bigcup_{n=1}^{\infty} B_{1/n}$

And together with A we find that

$A \cup B = \{\, x \, | \, f(x) \, \neq \, F(x) \, \}$
But $A \cup B$ is enumerable, so $[0,1] \setminus (A \cup B) \neq \emptyset$ , so there exists a point $x_0 \in [0,1]$ such that f(x_0) = F(x_0)

is a continuity point for

.

QED

Test your Math skills

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