Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals

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Theorem

Let $I$ be a real interval.

Let $x \in I$.

Let $f : I \to \R$ be a real function.

Then $f$ is right-continuous at $x$ if and only if:

for each real sequence $\sequence {x_n}_{n \mathop \in \N}$ such that $x_n \to x$ with $x_n > x$ for each $n$ we have:

$\map f {x_n} \to \map f x$

Corollary

Let $I$ be a real interval.

Let $x \in I$.

Let $f : I \to \R$ be a real function.

Then $f$ is right-continuous at $x$ if and only if:

for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:

$\map f {x_n} \to \map f x$

Proof

Necessary Condition

Suppose that is right-continuous at $x$.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence such that $x_n \to x$, with $x_n > x$ for each $n$.

We show that $\map f {x_n} \to \map f x$.

Let $\epsilon > 0$.

Since $f$ is right-continuous at $x$, there exists $\delta > 0$ such that whenever $x < y < x + \delta$, we have:

$\size {\map f x - \map f y} < \epsilon$

Since $x_n \to x$, there exists $N \in \N$ such that:

$\size {x_n - x} < \delta$

whenever $n \ge N$.

Since $x_n > x$, this is equivalent to:

$x < x_n < x + \delta$

So we have:

$\size {\map f x - \map f {x_n} } < \epsilon$

for $n \ge N$.

So:

$\map f {x_n} \to \map f x$

$\Box$

Sufficient Condition

We prove the contrapositive.

Suppose that $f$ is not right-continuous at $x$.

We show that:

there exists a real sequence $\sequence {x_n}_{n \mathop \in \N}$ such that $x_n \to x$ with $x_n > x$ for each $n$ but $\sequence {\map f {x_n} }_{n \mathop \in \N}$ does not converge to $\map f x$.

Since $f$ is not right-continuous at $x$, there exists some $\epsilon > 0$ such that for all $\delta > 0$:

there exists $y \in I$ with $x < y < x + \delta$ such that $\size {\map f x - \map f y} \ge \epsilon$

For each $n \in \N$ pick $x_n$ so that:

$\ds x < x_n < x + \frac 1 n$

and:

$\size {\map f x - \map f {x_n} } \ge \epsilon$

Then from the Squeeze Theorem, we have:

$x_n \to x$

while $x_n > x$ for each $n \in \N$.

However, then there exists no $N \in \N$ such that:

$\size {\map f x - \map f {x_n} } \ge \epsilon$

for all $n \ge N$, so:

$\sequence {\map f {x_n} }_{n \mathop \in \N}$ does not converge to $\map f x$.

$\blacksquare$