TITLE: Prove that the correspondence is lower hemicontinuous
QUESTION [0 upvotes]: Define the correspondence $\Phi : \mathbb R \to \mathbb R$ as $\Phi(x) = [x, x + 1]$. Prove that $\Phi$
is lower hemi-continuous.
I am struggling with how to prove this. Help would be greatly appreciated.

REPLY [0 votes]: I use the following definition of lower hemicontinuity: a correspondence $\Phi(x)$ is lower hemicontinuous if for any point $\{x,y\}$ in the graph of $\phi(x)$, and any sequence $x_{n}\rightarrow x$, there exists a sequence $y_{n}$ such that $y_{n}\rightarrow y$, and $y_{n}\in \Phi(x_{n})$ for all $n$.
On to your problem. Take $\{x,y\}$ in the graph of $\Phi(x)$ (this means that $y\in [x, x+1]$). Consider an arbitrary $x_{n}\rightarrow x$. Note that, for each $n$, $[x_{n}, x_{n}+1]$ is non-empty. Also, since $x_{n}\rightarrow x$ and $y\in [x, x+1]$, for large enough $n$ it must be the case that $y\in [x_{n}, x_{n}+1]$. Say that this is the case for all $n>N_{1}$.
Now, construct the $y_{n}$ sequence as follows. For all $n>N_1$, set $y_{n}$ equal to $y$. For all $n \leq N_{1}$, set $y_{n}$ to any element in $[x_{n}, x_{n+1}]$ (by the non-emptiness of these sets, we know that there is always something to choose).
Thus, we have a sequence, $y_{n}$, such that $y_{n}\rightarrow y$, and $y_{n}\in \Phi(x_{n})$, for all $n$.
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