# Thread: open and closed set

1. ## open and closed set

how does one go about showing that a set is open/closed.

e.g the set below is said tobe closed for finite values of T. but not closed for T= $\infty$
how can we show that?
$S= \{f(x):\int_0^T f(x) dx = 1\}$

2. Originally Posted by mtmath
how does one go about showing that a set is open/closed.

e.g the set below is said tobe closed for finite values of T. but not closed for T= $\infty$
how can we show that?
$S= \{f(x):\int_0^T f(x) dx = 1\}$
First, you need to be perfectly clear about the topological space in which this whole problem statement is to be situated.

If it is a metric space you can do the following:

1. To show that $S$ is closed (for $T\in \field{R}$) you show that the limit $f$ of any convergent sequence of functions $f_n\in S$ also lies in $S$.

2. To show that $S$ is not closed (for $T=\infty$) you show that there exists a sequence of functions $f_n$ in $S$, the limit $f$ of which does not lie in $S$. (For example: it is easy to find a sequence of functions $f_n\in S$ with limit $f \equiv 0,\notin S$.)

3. I'm not sure what the problem is asking for specifically. In general, to show a set is closed, you want to show its complement is open. Are you asking whether and why the set $S = \{x : \int_0^T f = 1\}$, specifically, is closed? I assume so. (Give enough detail with your problem!)

I suppose the integral is the usual Riemann integral and $f$ is Riemann-integrable on every compact subset of $\mathbb{R}$. Then $F(t)=\int_0^T f$ is continuous, and for any continuous function $F: \mathbb{R}\to \mathbb{R}$, and any $\alpha \in \mathbb{R}$, the set $\{x : F(x)=\alpha\}$ is closed. (Prove this! Just apply the definition of continuity.)

4. hi,
the set S is defined in the $L^2$ space of square-integrable functions.