Prove that intervals $(0,1)$ and $[0,1]$ are not homeomorphic without using connectedness properties

Question

I am trying to prove this statement, but having troubles with using notion of continuity.

The closest answer I found on stackexchange that matches the question is from Tarc (last answer):

Prove that an open interval and a closed interval are not homeomorphic

Still, I don't quite comprehend the last paragraph:

Suppose $f: [a,b] \to (c, d)$ is a homeomorphism. Observe that $f(a)\not=f(b)$ as $f$ is injective and consider the point $f(a)+f(b)\over 2$ at half the distance between $f(a)$ and $f(b)$. As $f$ is surjective, $f(x) = {f(a)+f(b)\over 2 }$ for some $x \in [a,b]$.

Now let $\delta$ be enough for both the distances between $f(x)$ and $f(a)$, and between $f(x)$ and $f(b)$ to be less than $\delta$ and yet the open interval centered at $f(x)$ of radius $\delta$ not to cover the whole $(c,d)$.

As $f$ is continuous, there must be an open interval centered at $x$ large enough to contain both $a$ and $b$ and whose image under $f$ is within a distance of $\delta$ from $f(x)$. As such an interval must ecompass the whole $[a,b]$, the function $f$ cannot be surjective for $\delta$ is chosen in such a way that there're points in $(c,d)$ with a distance from $f(x)$ greater than $\delta$.

P.S. Without using connectedness properties -- that means no intermediate value theorem either.

UPDATE. Background of the task: This is one of the exercises from book on differential topology. Author didn't introduce notions of compactness or connectedness yet. So that's why I am wondering how to prove the claim w/o using them.

Answer 1

Well, if they were homemorphic then there is a continuous bijection $\;[0,1]\to (0,1)\;$ , which would make $\;(0,1)\;$ compact just as $\;[0,1]\;$ is...

Answer 2

Compactness of $[0,1]$ is the easiest way. Or use the derived fact that a continuous function on $[0,1]$ has a maximum and a minumum. If the minimum is $m \in (0,1)$ for a continuous $f:[0,1] \to (0,1)$, then $f$ cannot be onto, as the values in $(0,m)$ are not assumed by $f$.

Answer 3

Note: I need new glasses (or my mind is going)! I answered the wrong question. I thought one interval was half-open and half-closed. But no harm, the arguments are very similar.

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A student has completed his study of Infima and suprema of real numbers and then learns the definitions for a topological space and continuous functions. He understands that

$\tag 1 f(\overline{A})\subseteq \overline{f(A)}$

whenever $f$ is continuous.

The student also learns about the two point discrete space $D = \{0,1\}$, and then works out the details for the following:

Proposition 1: There is no continuous surjective function $f: \mathbb R \to D$; in fact, there is no continuous surection of $\mathbb R$ onto any finite discrete space.
Proof
There is an open interval about the number $0$ such that all the numbers in that interval get mapped to the same point $f(0)$. Assume that there are real numbers that do not get mapped to $f(0)$. But then (a non-empty closed set of real numbers bounded from, say below, has a least element), there is a continuous function

$\tag 2 g: (a, b]: \to D \; \text{ with } a \lt 0 \lt b$ such that
$\quad g \text{ is constant and equal to } f(0) \text{ on the open interval } (a, b)$
$\quad g(b) \ne f(0)$

But this is absurd by (1). $\quad \blacksquare$

After some help from a Calculus Professor, he can claim without apology that the open interval $(0,1)$ is homeomorpihc to $\mathbb R$.

She then proves, without undue difficulty nor reading chapter 2 of his topology book, that $(0,1)$ can't be homeomorpic $(0,1]$.

Logic: If you remove any point from $(0,1)$ the result is the disjoint topological sum,

$\tag 3 \mathbb R \sqcup \mathbb R$

If you remove $\{1\}$ from $(0,1]$ the result is $\mathbb R$.

If the intervals are homeomorphic, then both $\mathbb R$ and $\mathbb R \sqcup \mathbb R$ must be, but that is absurd.

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Now we answer the stated question...

If you remove any one point from the open interval $(0,1)$ and remove another single point from the resulting space, you are looking at, up to a homeomorphism,

$\tag 4 \mathbb R \sqcup \mathbb R \sqcup \mathbb R$

But if you first take out $\{0\}$ from $[0,1]$ and then $\{1\}$, you are looking at $\mathbb R$. So the open and closed interval can't be homeomorphic.