Is it true that the cardinality of the range of any non-constant continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ is equal to cardinality of the set of real numbers?
If so, how do you prove it?
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Hint: Does the image necessarily contain an interval? If so, the answer is yes. – Mathematician 42 Apr 09 '17 at 13:24
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Yes. Is there a way to prove it? – Jus Apr 09 '17 at 13:26
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There is, I'll write it as an answer. – Mathematician 42 Apr 09 '17 at 13:26
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1Hint: If $f \colon \mathbb R \to \mathbb R$ is continuous, then it is uniquely determined by $f \restriction \mathbb Q$. – Stefan Mesken Apr 09 '17 at 13:27
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@AsafKaragila: It's not a duplicate of that question also Stefan's hint is more applicable to the other question. – Mathematician 42 Apr 09 '17 at 14:26
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@Mathematician42: Yes, you're right. Sorry about that. – Asaf Karagila Apr 09 '17 at 14:27
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@AsafKaragila: No problem, it might be a duplicate of another though but I couldn't find it immediately. – Mathematician 42 Apr 09 '17 at 14:28
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Possible duplicate of A continuous nonconstant function has uncountable range – martin.koeberl Apr 09 '17 at 16:13
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Although I think the answer here is better/more detailed. Maybe somebody else knows how to deal with that. – martin.koeberl Apr 09 '17 at 16:14
1 Answers
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Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a non-constant continuous function. Then there exist numbers $a,b\in \mathbb{R}$ such that $a\neq b$, $f(x)=a$ and $f(y)=b$ for some $x,y\in \mathbb{R}$. We may assume that $a<b$. By the intermediate value theorem, the interval $[a,b]$ belongs to the image.
Hence $(a,b)\subset \text{Im}(f)\subset \mathbb{R}$. It follows that $|(a,b)|\leq |\text{Im}(f)|\leq |\mathbb{R}|$, where $|A|$ denotes the cardinality of $A$. Since $|(a,b)|=|\mathbb{R}|$ we must have that $|\text{Im}(f)|=|\mathbb{R}|$.
To see that $|(a,b)|=|\mathbb{R}|$, first show that any two finite length intervals are bijective, then $|(a,b)|=|(-\frac{\pi}{2},\frac{\pi}{2})|$ and $\tan:(-\frac{\pi}{2},\frac{\pi}{2})\rightarrow \mathbb{R}$ is a bijection.
Mathematician 42
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Mathematician 42, can the same proof be applied to open intervals? Like (a,b), (a,b], [a,b) – Jus Apr 09 '17 at 13:32
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But from that how do we know that there is a bijection to the real line to show that the cardinality is equal to that of the real lines? – Jus Apr 09 '17 at 13:39
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Can I duplicate the above proof if i were to change the question from continuous functions to polynomial functions? Since polynomials are also continuous – Jus Apr 09 '17 at 14:05
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Yes, you can!!! (I just needed three exclamation marks to get the minimal length). – Mathematician 42 Apr 09 '17 at 14:23
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@Starfall: It answers the question asked by the OP, the linked question is a different question. – Mathematician 42 Apr 09 '17 at 14:29
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Exactly. three. (now I needed the .'s to get to 15 characters) – Mathematician 42 Apr 09 '17 at 14:55
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What about the set of integrable functions? If A is the set of all integrable functions, is the card of A still equal to that of real? – Jus Apr 09 '17 at 16:18
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That's a completely different question. You asked about the cardinality of the image of certain functions, now you are asking about the cardinality of a set of functions, that's very different. You should ask a new question if you're interested in that, be sure to take a look at the linked question first. – Mathematician 42 Apr 09 '17 at 16:22