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

In summary, the conversation discusses a proof involving showing that a function is continuous if its image is compact. The first direction is easily proven using basic facts, but the second direction involves looking at the graph of the function and proving it is homeomorphic to [0,1]. The distinction between the graph and image of the function is important in this proof.
  • #1
dumbQuestion
125
0
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.
 
Physics news on Phys.org
  • #2
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
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
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
dumbQuestion said:
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
oh i see. ok this will change things entirely. thank you so much for pointing this out
 

1. What is a compact graph?

A compact graph is a graph that can be contained within a finite space without any missing points. It is a topological property that means the graph has a finite number of points and is closed and bounded.

2. How is the compactness of a graph related to continuity?

In the context of functions, a graph is compact if and only if the function is continuous. This means that for every point on the graph, there is a corresponding point on the x-axis and the function approaches that point as the input approaches the given point. In other words, a continuous function has no gaps or jumps in its graph, making it compact.

3. What is the significance of a compact graph in mathematics?

A compact graph is important in mathematics because it allows us to generalize and prove certain theorems and properties. For example, a continuous function on a compact interval is guaranteed to have a maximum and minimum value. In addition, compactness is often used as a tool for proving more complex concepts in topology and analysis.

4. Can a discontinuous function have a compact graph?

No, a discontinuous function cannot have a compact graph. This is because a discontinuous function has at least one point where the function is not defined or has a jump. In this case, there would be a gap or missing point in the graph, making it non-compact.

5. How can we determine if a graph is compact or not?

To determine if a graph is compact, we can use the Heine-Borel theorem, which states that a subset of the real line is compact if and only if it is closed and bounded. This means that for a graph to be compact, it must have a finite number of points and all the points must be contained within a finite interval on the x-axis.

Similar threads

I Create a surjective function from [0,1]^n→S^n
    • Topology and Analysis
Replies
8
Views
1K
I Is [0,1] under the subspace topology Hausdorff, Compact, or Connected?
    • Topology and Analysis
Replies
15
Views
2K
I Why is [0,1] compact in the case of a collection of open neighborhoods?
    • Topology and Analysis
Replies
9
Views
3K
I Is the Set of Integer Outputs of sin(x) Sequentially Compact in ℝ?
    • Topology and Analysis
Replies
3
Views
844
I Understanding Compactness in ##R^2##: Analyzing Boundaries and Open Disks
    • Topology and Analysis
Replies
12
Views
2K
I Noncompact locally compact Hausdorff continuous mapping
    • Topology and Analysis
Replies
14
Views
2K
I Collection of finite unions of half-open intervals form an algebra
    • Topology and Analysis
Replies
3
Views
188
I Heine-Borel Theorem shouldn't work for open intervals?
    • Topology and Analysis
Replies
9
Views
2K
I Why the space X=(0,1) is (not sequentially) compact?
    • Topology and Analysis
Replies
7
Views
3K
Continuous function from (0,1) onto [0,1]?
    • Topology and Analysis
Replies
4
Views
14K

Hot Threads

Recent Insights

Back
Top