# Sufficient conditions for when continuity at a point implies continuity in a neighbourhood of that point

I consider a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ and I know that function $f$ is continuous at some $x_0=(y_0,z_0)$. Are there any other properties of function $f$ that can help me to conclude that function $f$ is in fact continuous in a neighborhood of $x_0$? I know, e.g., that function $f(y,z)$ is continuous from the right in $y$ for any $z$ and is continuous from the right in $z$ for any $y$. I also know that $f$ is non-decreasing in each component.

I am not interested in obvious properties that immediately imply the continuity of $f$ everywhere (such as differentiability, etc.).

I would argue that there is no such condition that doesn't feel unnecessarily strong.

Here are some examples:

1. The function defined by $$f(y,z) = \begin{cases} 0,&z\in\mathbb Q\\ y,&z\notin\mathbb Q \end{cases}$$ is continuous at the origin and continuous in $y$ for any fixed $z$, but it is not continuous in any neighborhood of the origin. This does not satisfy your assumptions yet, but it is still a good example to keep in mind.

2. The function defined by $$f(y,z) = \begin{cases} 0,&y\geq0\\ z,&y<0 \end{cases}$$ is continuous from the right in $y$ for any fixed $z$ and vice versa and continuous at the origin, but not continuous in any neighborhood of the origin.

3. You can make the previous example non-decreasing in both variables: $$f(y,z) = \begin{cases} 0,&y\geq0\text{ or }z\geq0\\ z,&y<0\text{ and }z<0 \end{cases}$$

• Thanks. I am aware of such an example. But I was hoping that those extra one-sided continuity properties that I have might help. The example you gave does not satisfy those. I also added monotonicity properties that I have. – Alik Mar 26 '18 at 14:50
• @Alik Sorry! I added another example. It's continuous from the right in both variables. Is that more satisfying? – Joonas Ilmavirta Mar 26 '18 at 14:55
• Yes, thank you. – Alik Mar 26 '18 at 14:58
• @Alik Turns out there is a counter-example which is non-decreasing in both arguments. – Joonas Ilmavirta Mar 26 '18 at 15:36
• Yes, thank you very much for all your help! – Alik Mar 26 '18 at 16:56