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

  • #1
125
0

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.
 

Answers and Replies

  • #2
22,097
3,283
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
WannabeNewton
Science Advisor
5,803
530
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
125
0
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
22,097
3,283
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.
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
125
0
oh i see. ok this will change things entirely. thank you so much for pointing this out
 

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

Replies
4
Views
10K
  • Last Post
2
Replies
35
Views
29K
Replies
3
Views
4K
Replies
7
Views
2K
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
9
Views
2K
Replies
7
Views
2K
Replies
2
Views
2K
Replies
2
Views
6K
Top