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

## Main Question or Discussion Point

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.

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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?

WannabeNewton