# Graph of f:[0,1]->R compact <=> f continuous

1. Jul 6, 2013

### dumbQuestion

I know this proof is probably super easy but I'm really stuck. I don't want someone to solve it for me, I just want a hint.

One way is trivial:

suppose f continuous.
[0,1] compact and the continuous image of a compact space is compact so f([0,1]) is compact

Now the other direction...

here's some basic facts i have to work with:

1) R is metrizable
2) R is hausdorf
3) R is regular
4) R is normal
5) f([0,1]) is compact (by assumption) (so it follows that every open cover has a finite subcover)
6) f([0,1]) is closed (follows from 5 because R is Hausdorf and compact subset of Hausdorf space is closed)
7) R is NOT compact, but it's connected
8) [0,1] is connected and compact (every open cover has finite subcover, and no separate of [0,1])

I don't know, everything I try I Just get no where. Can someone give a hint on how/where to start? In showing f is continuous should I try the approach 1) pre-image under f of every open set U in f([0,1]) is open in [0,1] ? 2) pre-image under f of every closed set U in f([0,1]) is closed in [0,1] ? 3) maybe try a contradiction? Assume f is not continuous, which means there exists an open set U in f([0,1]) such that it's pre image is not open, etc.

2. Jul 6, 2013

### micromass

Staff Emeritus
So let $G(f) = \{(x,f(x))~\vert~x\in [0,1]\}$. Maybe you can start by proving that $[0,1]$ and $G(f)$ are homeomorphic?

3. Jul 6, 2013

### WannabeNewton

dumbQuestion, your thread title says graph but in your main body it seems you are trying to show that if $f([0,1])$ is compact then $f$ is continuous but this is the image of $f$, not the graph of $f$, and it isn't true in general. Consider for example $f:[0,1] \rightarrow \mathbb{R}$ given by $f(x) = 0, 0 \leq x\leq \frac{1}{2}$ and $f(x) = 1, \frac{1}{2} < x\leq 1$. The image of this is $\{0,1\}$ which is compact but the function isn't continuous. So are you sure you aren't supposed to actually look at the graph of $f$?

4. Jul 6, 2013

### dumbQuestion

oh yes, I am supposed to look at the graph of f. I guess I was just mistaking the graph of f for the image of f. I guess I'm not seein gthe distinction between the graph and image.

5. Jul 6, 2013

### micromass

Staff Emeritus
The graph of $f$ is by definition a subset of $[0,1]\times \mathbb{R}$. While the image is a subset of $\mathbb{R}$. They are two very different different things.

6. Jul 6, 2013

### dumbQuestion

oh i see. ok this will change things entirely. thank you so much for pointing this out