Prove that a function is sequentially lower semicontinuous

Question

Let be $(X, \{ p_i \}_{i \in I} )$ a locally convex space, $M_0\subset X$ a bounded and nonempty set and $f = l + I_{M_0}$ where l is a continuous function and \begin{equation*} I_{M_0}(x)= \begin{cases} 1 & \text{if} \hspace{.1cm} x \in M_0 \\ 0 & \text{if} \hspace{.1cm} x \notin M_0 \end{cases} \end{equation*} Then f is a sequentially lower semicontinuous function? I guess the problem comes down to prove that $I_{M_0}$ is sequentially lower semicontinuous since $l$ is a continuos function. Update: $M_0$ is sequentially closed.

Answer 1

The question has been updated in a comment.

False even on the real line. Can you think of a bounded non-empty set $M_0$ such that $I_{M_0}$ is not l.s.c.? Hint: take $M_0=(0,1]$.

Answer to the edited question: again consider the real line. if $M_0$ is closed then $I_{M_0}$ is upper semicontinuous, not lower semicontinuous. For example take $M_0=[0,1]$ and consider the sequence $x_n =1+\frac 1 n$. This sequence tends to $x=1$, $I_{M_0}(x_n)=0$ for all $n$, $I_{M_0}(x)=1$. Hence $I_{M_0}$ is not lower semicontinuous.