Prove ceiling function is lower semicontinuous

Question

Can someone give me a hand with this exercise please? I want to prove that the ceiling function is lower semicontinuous, but I am not sure how to do it. It comes in my book as an example, right after the definition, without a proof, so I guess it is very easy. I understand why it is so intuitively but when I try to apply the definition I can't find the delta.

Just for the record, my definition of l.s.c:

$f: X\to\mathbb R$ is lower semicontinuous on $X$ if for all $c< f(x)$ there exists $d>0$ such that $c< f(y)$ whenever $\|x,y\|< d$

And the ceiling function is the function such that $f(t)=n$ if $n-1< t<= n$

Answer 1

You start by going through the definition.

First, the definition says "for all $x$ and $c$ such that $c<f(x)$", which means that your proof must start with

Let $x,c$ be such that $c<f(x)$

Which in your case will be

Let $x,c$ be such that $c<\lceil x \rceil$.

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Now the next part of the definition says "there exists such a $d$..." which means you have to do some work. And that also means I will, at first, only provide hints because they are more useful than full results.

Specifically, you need to find such a $d$ that will force $f(y)$ (in your case, $\lceil y\rceil$) to also be greater than $c$.

You do that by thinking about $x,y$ and $f$.

• Think about what the value of $f(x)$ is.
• Now think about the value of $f(y)$ when $y$ is near $x$. For example, if $x=\frac12$, what is the value of $f(y)$ if $|x-y|<\frac14$?
• What about if $x=1.0001$? What do you have to set $d$ to in order to get the same behaviour as above?