Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals/Corollary
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 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 $f$ is continuous at $x$, then:
- for each real sequence $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, converging to $x$ we have:
- $\map f {x_n} \to \map f x$
from Sequential Continuity is Equivalent to Continuity in the Reals.
So in particular:
- for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
- $\map f {x_n} \to \map f x$
$\Box$
Sufficient Condition
Suppose that:
- for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
- $\map f {x_n} \to \map f x$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence, with $x_n > x$ for each $n$, converging to $x$.
Let $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ be a subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$.
We aim to show that $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ has a subsequence converging to $\map f x$.
We will then obtain $\map f {x_n} \to \map f x$ from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit.
From the Peak Point Lemma, $\sequence {x_{n_j} }_{j \mathop \in \N}$ has a monotonic subsequence $\sequence {x_{n_{j_k} } }_{k \mathop \in \N}$.
Since $x_n > x$ for each $n$, we must have $x_{n_{j_k} } > x$ for each $k$.
So from hypothesis:
- $\map f {x_{n_{j_k} } } \to \map f x$
So:
- $\sequence {\map f {x_{n_{j_k} } } }_{j \mathop \in \N}$ is a subsequence of $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ converging to $\map f x$.
Since $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ was an arbitrary subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$, we have:
- $\map f {x_n} \to \map f x$
from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit.
Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary, we have:
- for each real sequence $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, converging to $x$ we have $\map f {x_n} \to \map f x$.
So, from Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals, we have:
- $f$ is right-continuous at $x$.
$\blacksquare$