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Some T/F questions. Instead of doing strict proofy questions, I am trying to understand the topic and making sure whether I am clear on the topic. Let me know whether I am right or wrong and I'll appreciate if you briefly explain:) Thanks for helping!

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For number three:

a. Yes

b. No, For example, f(x) as a constant function?

c. Yes, that's another definition of compactness

d. No, for similar reason above

e. Yes, an alternative definition of compactness

For number four:

  1. a. yes b. yes c. No

AM i correct?

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2 Answers 2

up vote 1 down vote accepted

Your answer for 3. b) seems wrong to me. If $X$ is compact then $f(X)$ is compact. Therefore $f(X)$ is closed in $\mathbb{R}^n$. Since $f$ is continuous, $f(X)$ is closed.

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In your answer, you mentionned the case of a constant function. Let's say that $f(x) = c$ for all $x \in \mathbb{R}^n$. Then, either $c \not\in X$ and $f^{-1}(X) = \emptyset$ or $c \in X$ and $f^{-1}(X) = \mathbb{R}^n$ which is closed. –  M.G Dec 20 '12 at 1:17
    
I don't follow your argument; in particular the last sentence. Why not simply state "$X$ is compact, hence closed. So, $f^{-1}(X)$ is closed, since $f$ is continuous."? –  David Mitra Dec 20 '12 at 1:59

As M.G has already pointed out, your answer to the second part of (3) is wrong, but the rest of (3) is correct. (4) has four parts, and you’ve provided only three answers, so I’m not sure which are right. I’ll go through all four parts.

(a) If $n\ge 2$ there are connected sets in $\Bbb R^n$ that are not path connected; an example in $\Bbb R^2$ is the topologist’s sine curve.

(b) $f^{-1}[X]$ need not be connected. For example, if $f:\Bbb R\to\Bbb R:x\mapsto x^2$, and $X=\{1\}$, then $f^{-1}[X]=\{-1,1\}$, which is not connected.

(c) However, continuous functions preserve connectedness, so yes, $f[X]$ must be connected.

(d) Yes: this is one definition of connectedness.

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