I am teaching Mathematical analysis. A student asked this question. I think this is a good question, but don't know the answer.



Is there a function defined on real numbers which is continuous from the left, but not from the right, everywhere
Here's an elementary construction that may be more digestable by your student. 


1. By definition of left continuity, for every point $x$ and every $\epsilon > 0$ there exists $\delta > 0$ such that if $y\in (x-\delta,x)$, $|f(y)-f(x)| < \epsilon$. We can upgrade this by the triangle inequality to for every $y\_1,y\_2\in (x-\delta,x)$, $|f(y\_1) - f(y\_2)| < 2 \epsilon$.
2. Now we construct a sequence of points $x\_i$ as follows. Let $\epsilon\_i = 2^{-i}$ be fixed. Let $x\_1$ be arbitrary, and $\delta\_0 = 1$. Given $x\_k$, we define $\tilde{\delta}\_k$ be that which is given by step 1. And we define $$\delta\_k = \min(\tilde{\delta}\_k, \frac13 \delta\_{k-1})$$
And we let $x\_{k+1} = x\_k - \frac13 \delta\_k$.
3. Let $z = \lim x\_k$. Note that $z < x\_k$ for all $k$ and $\frac13 \delta\_k < x\_k - z < \frac23 \delta\_k$, so by construction we have that whenever $y \in (z - \frac13 \delta\_k, z+\frac13\delta\_k)$ we have $|f(x) - f(y)| < 2^{1-k}$. This shows continuity of $f$ at $x$.


