# Does there exist a function $f:\mathbb{R} \to \mathbb{R}$ which is 'strongly discontinuous' at some point?

A function $f:\mathbb{R} \to \mathbb{R}$ is discontinuous at $x_0 \in \mathbb{R}$ if there exists a sequence $a_n$ such that $a_n \to x_0$ but $f(a_n) \not \to f(x_0)$.

I thought of a way of strengthening this by saying that a function $f:\mathbb{R} \to \mathbb{R}$ is strongly discontinuous at $x_0 \in \mathbb{R}$ if for every injective sequence $a_n \to x_0$, $f(a_n)$ diverges.

But I can't think of an example of a function with this property. Does such a function exist?

• $f(x) = 1/x$ with $f(0)=0$? – bitesizebo Apr 3 '18 at 3:32
• By injective sequence do you mean a sequence all of whose terms are distinct? – carmichael561 Apr 3 '18 at 3:33
• @Z.Xie Ah, of course. That works. Is there an example where each sequence $f(a_n)$ doesn't converge at all (i.e., not even to infinity), or alternatively an example where $f$ is bounded? – MathematicsStudent1122 Apr 3 '18 at 3:35
• If bounded we could always apply Bolzano Weierstrass – bitesizebo Apr 3 '18 at 3:41
• @Z.Xie Right, perfect. Thanks. – MathematicsStudent1122 Apr 3 '18 at 3:43

Maybe I'm misinterpreting this, but doesn't $f(x)=1/x$ and $x_0 = 0$ work? Any sequence tending towards zero will correspondingly tend towards infinity.

• This answers the Q as it is stated. In the comments the proposer asks whether it is possible for all $x_0\in \Bbb R,$ which is a different Q, to which I have given an answer.....+1 – DanielWainfleet Apr 3 '18 at 17:20

The Q as written has been answered by ASKASK. The A to the Q in the proposer's comments, which is whether it can hold for every $x_0\in \Bbb R$, is NO.

For $n\in \Bbb N$ let $S(n)=\{y\in \Bbb R: |f(y)|>n\}.$

Suppose that for all $x\in \Bbb R$ and for every $(a_n)_n$ sequence in $\Bbb R\setminus \{x\}$ converging to $x,$ the sequence $(f(a_n))_n$ does not converge.

Then for $x\in \Bbb R$ and $n\in \Bbb N$ there must exist $r_{x,n}>0$ such that $$A(x,n)=(-r_{x,n}+x,\;r_{x,n}+x)\setminus \{x\} \subset S(n).$$ Otherwise $\forall s\in \Bbb N \;\exists a_s\; (0<|a_s-x|<1/s \land |f(a_s)|\leq n)$ but then $(f(a_s))_s$ has a convergent sub-sequence.

Let $U(n)=\cup_{x\in \Bbb R}A(x,n).$ Then $U(n)$ is open and dense.

The Baire Category Theorem implies that $\cap_{n\in \Bbb N}U(n)\ne \emptyset.$ But if $z\in U(n)$ for every $n$ then $z\in S(n)$ for every $n,$ so $|f(z)|>n$ for every $n\in \Bbb N,$ which is absurd.