# Can a function have uncountably many jump discontinuities? [duplicate]

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I understand the proof that shows that if $f$ is monotone on an interval, then it has at most countably many jump discontinuities. I feel like a similar idea also allows us to show that any function can have at most countably many jump discontinuities. Here is my attempt at a proof.

If a jump continuity of $f$ exists at $x$, this means that $f(x^-)$ and $f(x^+)$ exist, but $f(x^-)\neq f(x^+)$. The existence of left and right hand limits means that for any $\epsilon > 0$, we can find a $\delta > 0$ such that if $y\in (x,x+\delta)$, then $|f(x^+) - f(y)| < \epsilon$ and if $y\in (x-\delta, x)$, then $|f(x^-)-f(y)| < \epsilon$. Thus, the only discontinuity (jump or otherwise) in the interval $(x-\delta, x+\delta)$ is at $x$.

I believe that this means that the jump discontinuities only exist within disjoint open intervals, of which there are at most countably many. If there does appear to be an issue at the boundary of the interval, for example, in a case where $x<y$ are both jump discontinuities and $x+\delta_x = y-\delta_y$, then we can make $\delta_x$ or $\delta_y$ a bit smaller to prevent this overlap.

Is this proof correct, or am I missing something? This statement seems a bit strong, especially since I've only heard a similar statement for monotone functions on an interval. If it's not true, can someone give a counterexample?

## marked as duplicate by José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 16 '18 at 17:38

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• Maybe I’m misunderstanding what a jump discontinuity is, but can’t we create a simple function with uncountably many discontinuities? Take $f(x) = x$ if $x \in \Bbb{Q}$, 0 otherwise. Doesn’t this function have uncountably many discontinuities, since there are uncountably many irrationals? – PhysMath Sep 16 '18 at 17:35
• The set of jump discontinuities of any function is countable, for the reasons you give. But a function can have an uncountable number of discontinuous points that aren't jump discontinuities. – TonyK Sep 16 '18 at 17:38
• @PhysMath Yes, you're misunderstanding what a jump discontinuity is. – David C. Ullrich Sep 16 '18 at 17:38
• The sentence about how there are no other discontinuiuties in that interval simply doesn't follow. In fact there can be infinitely many other jump discontinuities in that interval, even for a monotone function. What's true is that there cannot be any other jump continuitites with the "size" of the jump at least $\epsilon$ in that interval; as you see from the answer in the duplicate question, this shows that yes, the set of jump discontinuiuites is countable. – David C. Ullrich Sep 16 '18 at 18:02
• @DavidC.Ullrich That makes a lot sense, thank you! – J. Pistachio Sep 16 '18 at 18:15

## 1 Answer

Your assertion that the only discontinuity in $(x-\delta, x+\delta)$ is at $x$ is incorrect. Consider for example the function $f$ such that $f(x)=x$ on $\mathbb{Q}$ and $f(x)=1-x$ on $\mathbb{R} \backslash \mathbb{Q}$. It is discontinuous everywhere except $x=1/2$.

I'm not sure whether your assertion is correct, though.

• But those are noot jump discontinuiuties. – David C. Ullrich Sep 16 '18 at 17:37
• Right, but OP's argument was that the existence of left and right limits of $x$ implied that there was a neighborhood of $x$ where $x$ is the only discontinuity. – Sambo Sep 16 '18 at 17:57