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I'm reading Intro to Topology by Mendelson.

The problem statement is in the title. Also, I'm in the section right before he introduces the Intermediate Value Theorem, so I want to try and not use it in the proof.

My attempt at the proof is,

Let $a,b\in\mathbb{R}$ and $\mathfrak{J}=\left\{(a,b),[a,b],[a,b),(a,b],(-\infty,a),(-\infty,a],(a,\infty),[a,\infty),(-\infty,\infty)\right\}$. There are two cases to consider. Consider first that $f$ is a constant function. Then for any $W\in\mathfrak{J}$, $f(W)$ is real number, call it $c$. Hence, $f(W)=c$. Now suppose that $f$ is a non-constant function. Then for any $W\in\mathfrak{J}$, $f(W)\subset\mathbb{R}$. Let $a,b\in f(W)$ with $a<b$. Since $f$ is continuous, for each $x$ such that $a<x<b$, we must have $x\in f(W)$. Thus, $f(W)$ is an interval.

I'm wondering if I'm approaching problem correctly by the cases I chose. Also, I'm a little concerned about whether or not the second before last sentence is valid.

Thanks for any feedback or hints!

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1  
"Let $a,b\in f(W)$ with $a<b$. Since $f$ is continuous, for each $x$ such that $a<x<b$, we must have $x\in f(W)$." This is exactly the statement of the IVT... –  Sharkos Jul 29 '13 at 11:20
    
Hm, well that kind of sucks. Cause I was trying to work off the definition of what it means to be an interval and intuitively felt that was true, hence my concern. I see now you're right I am using IVT. Is this really the only approach or is there another way to do this? –  Shant Danielian Jul 29 '13 at 11:22
3  
Was the concept of connectedness already treated? –  Daniel Fischer Jul 29 '13 at 11:25
    
Yes it was. I was just looking over the previous section and saw that a continuous mapping of $X$ onto $Y$, with $X$ connected means that $Y$ is also connected. I'm wondering, if I consider the mapping $f:W\to f(W)$, where $W\in\mathfrak{J}$, is that an onto function since it's continuous? –  Shant Danielian Jul 29 '13 at 11:28
1  
Yes, provided that you know (or show) that intervals are the only connected subspaces of $\mathbb{R}$ (I am including the singleton $\{a\}$ as the trivial interval $[a,a]$). –  Aleš Bizjak Jul 29 '13 at 12:04

2 Answers 2

up vote 0 down vote accepted

From the hints given in the comments I was able to complete the proof.

Let $a,b\in\mathbb{R}$ and $\mathfrak{J}=\left\{(a,b),[a,b],[a,b),(a,b],(-\infty,a),(-\infty,a],(a,\infty),[a,\infty),(-\infty,\infty)\right\}$. There are two cases to consider. Consider first that $f$ is a constant function. Then for any $W\in\mathfrak{J}$, $f(W)$ is real number, call it $c$. Hence, $f(W)=c.$ Now suppose that $f$ is a non-constant function. Then for any $W\in\mathfrak{J}$, $f(W)\subset\mathbb{R}$. We know that $f:W\to f(W)$ is continuous and onto and since $W$ is connected, we must have $f(W)$ connected and thus an interval.

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The only connected subsets of $\mathbb R$ are intervals, and continuous image of connected set is connected.

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"The only connected subsets of $\mathbb R$ are the intervals" is basically the IVT. –  Najib Idrissi Aug 5 '13 at 6:57
    
One can show that any subset of the real line which has two distinct points is connected if and only if it is an interval, without explicitly using the IVT. –  Shant Danielian Aug 5 '13 at 9:50

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