Slutsky's Theorem, Jun Shao's Mathematical Statistics

Question

The statement of the theorem is that if $X, X_1, X_2, \cdots, Y_1, Y_2, \cdots$ are random variables on a probability space, $X_n$ converges in distribution to $X$ and $Y_n$ converges to $c$ a constant in probability, then $X_n + Y_n$ converges to $X + c$ in distribution.

My only question about the proof is the claim that $\epsilon$ can be arbitrary. I understand that if $t - c$ is not a continuity points of $F_X$, then the statement holds trivially for $t$, so I only consider the case when $t - c$ is a continuity point. But how do I reason that $t - c + \epsilon$ and $t - c - \epsilon$ are continuity points of $F_X$ for arbitrary $\epsilon$, given that $t - c$ is a continuity point? (The inequalities don't need to hold if $t - c + \epsilon$ and $t - c - \epsilon$ are not continuity points, right?)

Thank you very much in advance!

Here's the statement and the proof:

Answer 1

The justification will use that well known result that set of discontinuity points of cdf is countable.

Take any $N \in \mathbb{N}$, any $x$ continuity point of $F$. Enough to show that we can choose some $\epsilon \in (0,\frac{1}{N})$ such that $x- \epsilon$ and $x + \epsilon$ are both continuity points of $F$.

$A_{N,x} := \{ \epsilon \in (0,\frac{1}{N}) | x - \epsilon \text{ is cont. pt.} \}$

$B_{N,x} := \{ \epsilon \in (0,\frac{1}{N}) | x + \epsilon \text{ is cont. pt.} \}$

Enough to show that $A_{N,x} \cap B_{N,x}$ is nonempty. Suppose not. Then, $A_{N,x} \subseteq B_{N,x}^c$. Now, $\epsilon \in A_{N,x} \implies (x+ \epsilon) \text{ is discontinuity point of F}$

But, $A_{N,x}$ has uncountably many members ( as set of discontinuity points is countable) . This means that there are uncountably many discontinuity points of $F$. Contradiction !